数列极差问题.docx
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数列极差问题
在黑板上写了N个正整数组成的一个数列,进行如下操作:
每次擦去其中的两个数a和b,然后在数列中加入一个数a×b+1,如此下去直至黑板上剩下一个数,在所有按这种操作方式最后得到的数中,最大的为max,最小的为min,则该数列的极差定义为M=max-min。
请你编程,对于给定的数列,计算极差。
输入
输入包含多个测试集。
每个测试集的第一个数N表示正整数序列长度(0<=N<=50000),随后是N个正整数。
N为0表示输入结束。
输出
每个结果一行
输入样例
3
1
2
0
输出样例
来源:
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Eejit 发布时间:
2011-5-2715:
14:
55
表示与sea313081574想的一样,不知道对不对
xuke 发布时间:
2011-5-2522:
42:
08
#include
#define maxn 999999999
using namespace std;
int ff(int n)
{ if(n==0) return 1;
int sum=0;
while(n)
{
sum++;
n=n/10;
}
return sum;
bool cmp(const int a,const int b)
{return a>b;
int main()
int a[2002];
int b[2002];
int n;
int temp;
int max,min;
int i;
while(cin>>n,n)
for(i=0 ;i { cin>>a[i]; b[i]=a[i]; } temp=n; while(temp>1) { sort(a,a+n); a[1]=a[0]*a[1]+1; a[0]=maxn; temp--; } max=a[1]; //cout< temp=n; while(temp>1) { sort(b,b+n,cmp); b[1]=b[0]*b[1]+1; b[0]=0; temp--; } min=b[1]; // cout< /////////////// // cout< cout< }return 0;}评论人:sea313081574 发布时间:2010-12-3010:09:29如有一组数字 a1则最大的就是(((a1*a2+1)*a3+1)*a4+1.........)*an+1最小的就是(((an*an-1+1)*an-2+1)*an-3+1.........)*a1那剩下来就简单了。评论人:mykeping 发布时间:2010-5-1722:39:43#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:mykeping 发布时间:2010-5-1722:39:34#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1519:07:44#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1516:14:48以下是极差概念,指一组数据中最大数据与最小数据的差,在统计中常用极差来刻画一组数据的离散程度。以及表示,R=Xmax-Xmin。又称全距或范围误差。反映的是变量分布的变异范围和离散幅度,在总体中任何两个单位的标准值之差都不能超过极差。同时,它能体现一组数据波动的范围。 如 12 12 13 14 16 21 这组数的极差就是 21-12=9与题中不同,但是题中结果似乎是这么的来的,请版主提示评论人:leijf2010 发布时间:2010-5-1421:54:01我不知道自己的解法对不对,因为我还有点不明白这个题目的意思,如果是随便擦出其中的两个数,那么最后那个最大值和最小值是不是也要算上擦去两个数的乘积加1? 如:1 2 3,如果擦去的是1,2,那么加入的就是3。如果擦去2,3,加入的值就是7,那么最大值和最小值之差就是6。而不是2?自己的一些想法,不知道对不对?#include#includeusing namespace std;#define N 50000vectorb;int main(){ int A[N]; int n,j,k,l,m,sum; int max,min; int*p; cin>>n; for(int i=0;i { if(cin>>A[i]) b.push_back(A[i]); } k=b.size()-1; max=b[k]; min=b[k]; for(j=b.size()-1;j>0;j--) { l=b[j]; m=b[j-1]; if(m>max) max=m; else min=m; sum=l*m+1; p=&b[j]; b.erase(p-1,p); b.push_back(sum); cout< } cout< return 0;}评论人:hanhan0121 发布时间:2010-3-723:54:41#includevoid main(){ int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t; scanf("%d",&N); t=N; for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
cin>>a[i];
b[i]=a[i];
temp=n;
while(temp>1)
{ sort(a,a+n);
a[1]=a[0]*a[1]+1;
a[0]=maxn;
temp--;
max=a[1];
//cout< temp=n; while(temp>1) { sort(b,b+n,cmp); b[1]=b[0]*b[1]+1; b[0]=0; temp--; } min=b[1]; // cout< /////////////// // cout< cout< }return 0;}评论人:sea313081574 发布时间:2010-12-3010:09:29如有一组数字 a1则最大的就是(((a1*a2+1)*a3+1)*a4+1.........)*an+1最小的就是(((an*an-1+1)*an-2+1)*an-3+1.........)*a1那剩下来就简单了。评论人:mykeping 发布时间:2010-5-1722:39:43#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:mykeping 发布时间:2010-5-1722:39:34#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1519:07:44#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1516:14:48以下是极差概念,指一组数据中最大数据与最小数据的差,在统计中常用极差来刻画一组数据的离散程度。以及表示,R=Xmax-Xmin。又称全距或范围误差。反映的是变量分布的变异范围和离散幅度,在总体中任何两个单位的标准值之差都不能超过极差。同时,它能体现一组数据波动的范围。 如 12 12 13 14 16 21 这组数的极差就是 21-12=9与题中不同,但是题中结果似乎是这么的来的,请版主提示评论人:leijf2010 发布时间:2010-5-1421:54:01我不知道自己的解法对不对,因为我还有点不明白这个题目的意思,如果是随便擦出其中的两个数,那么最后那个最大值和最小值是不是也要算上擦去两个数的乘积加1? 如:1 2 3,如果擦去的是1,2,那么加入的就是3。如果擦去2,3,加入的值就是7,那么最大值和最小值之差就是6。而不是2?自己的一些想法,不知道对不对?#include#includeusing namespace std;#define N 50000vectorb;int main(){ int A[N]; int n,j,k,l,m,sum; int max,min; int*p; cin>>n; for(int i=0;i { if(cin>>A[i]) b.push_back(A[i]); } k=b.size()-1; max=b[k]; min=b[k]; for(j=b.size()-1;j>0;j--) { l=b[j]; m=b[j-1]; if(m>max) max=m; else min=m; sum=l*m+1; p=&b[j]; b.erase(p-1,p); b.push_back(sum); cout< } cout< return 0;}评论人:hanhan0121 发布时间:2010-3-723:54:41#includevoid main(){ int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t; scanf("%d",&N); t=N; for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
sort(b,b+n,cmp);
b[1]=b[0]*b[1]+1;
b[0]=0;
min=b[1];
// cout< /////////////// // cout< cout< }return 0;}评论人:sea313081574 发布时间:2010-12-3010:09:29如有一组数字 a1则最大的就是(((a1*a2+1)*a3+1)*a4+1.........)*an+1最小的就是(((an*an-1+1)*an-2+1)*an-3+1.........)*a1那剩下来就简单了。评论人:mykeping 发布时间:2010-5-1722:39:43#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:mykeping 发布时间:2010-5-1722:39:34#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1519:07:44#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1516:14:48以下是极差概念,指一组数据中最大数据与最小数据的差,在统计中常用极差来刻画一组数据的离散程度。以及表示,R=Xmax-Xmin。又称全距或范围误差。反映的是变量分布的变异范围和离散幅度,在总体中任何两个单位的标准值之差都不能超过极差。同时,它能体现一组数据波动的范围。 如 12 12 13 14 16 21 这组数的极差就是 21-12=9与题中不同,但是题中结果似乎是这么的来的,请版主提示评论人:leijf2010 发布时间:2010-5-1421:54:01我不知道自己的解法对不对,因为我还有点不明白这个题目的意思,如果是随便擦出其中的两个数,那么最后那个最大值和最小值是不是也要算上擦去两个数的乘积加1? 如:1 2 3,如果擦去的是1,2,那么加入的就是3。如果擦去2,3,加入的值就是7,那么最大值和最小值之差就是6。而不是2?自己的一些想法,不知道对不对?#include#includeusing namespace std;#define N 50000vectorb;int main(){ int A[N]; int n,j,k,l,m,sum; int max,min; int*p; cin>>n; for(int i=0;i { if(cin>>A[i]) b.push_back(A[i]); } k=b.size()-1; max=b[k]; min=b[k]; for(j=b.size()-1;j>0;j--) { l=b[j]; m=b[j-1]; if(m>max) max=m; else min=m; sum=l*m+1; p=&b[j]; b.erase(p-1,p); b.push_back(sum); cout< } cout< return 0;}评论人:hanhan0121 发布时间:2010-3-723:54:41#includevoid main(){ int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t; scanf("%d",&N); t=N; for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
///////////////
// cout< cout< }return 0;}评论人:sea313081574 发布时间:2010-12-3010:09:29如有一组数字 a1则最大的就是(((a1*a2+1)*a3+1)*a4+1.........)*an+1最小的就是(((an*an-1+1)*an-2+1)*an-3+1.........)*a1那剩下来就简单了。评论人:mykeping 发布时间:2010-5-1722:39:43#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:mykeping 发布时间:2010-5-1722:39:34#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1519:07:44#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1516:14:48以下是极差概念,指一组数据中最大数据与最小数据的差,在统计中常用极差来刻画一组数据的离散程度。以及表示,R=Xmax-Xmin。又称全距或范围误差。反映的是变量分布的变异范围和离散幅度,在总体中任何两个单位的标准值之差都不能超过极差。同时,它能体现一组数据波动的范围。 如 12 12 13 14 16 21 这组数的极差就是 21-12=9与题中不同,但是题中结果似乎是这么的来的,请版主提示评论人:leijf2010 发布时间:2010-5-1421:54:01我不知道自己的解法对不对,因为我还有点不明白这个题目的意思,如果是随便擦出其中的两个数,那么最后那个最大值和最小值是不是也要算上擦去两个数的乘积加1? 如:1 2 3,如果擦去的是1,2,那么加入的就是3。如果擦去2,3,加入的值就是7,那么最大值和最小值之差就是6。而不是2?自己的一些想法,不知道对不对?#include#includeusing namespace std;#define N 50000vectorb;int main(){ int A[N]; int n,j,k,l,m,sum; int max,min; int*p; cin>>n; for(int i=0;i { if(cin>>A[i]) b.push_back(A[i]); } k=b.size()-1; max=b[k]; min=b[k]; for(j=b.size()-1;j>0;j--) { l=b[j]; m=b[j-1]; if(m>max) max=m; else min=m; sum=l*m+1; p=&b[j]; b.erase(p-1,p); b.push_back(sum); cout< } cout< return 0;}评论人:hanhan0121 发布时间:2010-3-723:54:41#includevoid main(){ int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t; scanf("%d",&N); t=N; for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
cout< }return 0;}评论人:sea313081574 发布时间:2010-12-3010:09:29如有一组数字 a1则最大的就是(((a1*a2+1)*a3+1)*a4+1.........)*an+1最小的就是(((an*an-1+1)*an-2+1)*an-3+1.........)*a1那剩下来就简单了。评论人:mykeping 发布时间:2010-5-1722:39:43#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:mykeping 发布时间:2010-5-1722:39:34#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1519:07:44#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1516:14:48以下是极差概念,指一组数据中最大数据与最小数据的差,在统计中常用极差来刻画一组数据的离散程度。以及表示,R=Xmax-Xmin。又称全距或范围误差。反映的是变量分布的变异范围和离散幅度,在总体中任何两个单位的标准值之差都不能超过极差。同时,它能体现一组数据波动的范围。 如 12 12 13 14 16 21 这组数的极差就是 21-12=9与题中不同,但是题中结果似乎是这么的来的,请版主提示评论人:leijf2010 发布时间:2010-5-1421:54:01我不知道自己的解法对不对,因为我还有点不明白这个题目的意思,如果是随便擦出其中的两个数,那么最后那个最大值和最小值是不是也要算上擦去两个数的乘积加1? 如:1 2 3,如果擦去的是1,2,那么加入的就是3。如果擦去2,3,加入的值就是7,那么最大值和最小值之差就是6。而不是2?自己的一些想法,不知道对不对?#include#includeusing namespace std;#define N 50000vectorb;int main(){ int A[N]; int n,j,k,l,m,sum; int max,min; int*p; cin>>n; for(int i=0;i { if(cin>>A[i]) b.push_back(A[i]); } k=b.size()-1; max=b[k]; min=b[k]; for(j=b.size()-1;j>0;j--) { l=b[j]; m=b[j-1]; if(m>max) max=m; else min=m; sum=l*m+1; p=&b[j]; b.erase(p-1,p); b.push_back(sum); cout< } cout< return 0;}评论人:hanhan0121 发布时间:2010-3-723:54:41#includevoid main(){ int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t; scanf("%d",&N); t=N; for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
return 0;
sea313081574 发布时间:
2010-12-3010:
09:
29
如有一组数字 a1则最大的就是(((a1*a2+1)*a3+1)*a4+1.........)*an+1最小的就是(((an*an-1+1)*an-2+1)*an-3+1.........)*a1那剩下来就简单了。评论人:mykeping 发布时间:2010-5-1722:39:43#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:mykeping 发布时间:2010-5-1722:39:34#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1519:07:44#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1516:14:48以下是极差概念,指一组数据中最大数据与最小数据的差,在统计中常用极差来刻画一组数据的离散程度。以及表示,R=Xmax-Xmin。又称全距或范围误差。反映的是变量分布的变异范围和离散幅度,在总体中任何两个单位的标准值之差都不能超过极差。同时,它能体现一组数据波动的范围。 如 12 12 13 14 16 21 这组数的极差就是 21-12=9与题中不同,但是题中结果似乎是这么的来的,请版主提示评论人:leijf2010 发布时间:2010-5-1421:54:01我不知道自己的解法对不对,因为我还有点不明白这个题目的意思,如果是随便擦出其中的两个数,那么最后那个最大值和最小值是不是也要算上擦去两个数的乘积加1? 如:1 2 3,如果擦去的是1,2,那么加入的就是3。如果擦去2,3,加入的值就是7,那么最大值和最小值之差就是6。而不是2?自己的一些想法,不知道对不对?#include#includeusing namespace std;#define N 50000vectorb;int main(){ int A[N]; int n,j,k,l,m,sum; int max,min; int*p; cin>>n; for(int i=0;i { if(cin>>A[i]) b.push_back(A[i]); } k=b.size()-1; max=b[k]; min=b[k]; for(j=b.size()-1;j>0;j--) { l=b[j]; m=b[j-1]; if(m>max) max=m; else min=m; sum=l*m+1; p=&b[j]; b.erase(p-1,p); b.push_back(sum); cout< } cout< return 0;}评论人:hanhan0121 发布时间:2010-3-723:54:41#includevoid main(){ int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t; scanf("%d",&N); t=N; for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
则最大的就是
(((a1*a2+1)*a3+1)*a4+1.........)*an+1
最小的就是
(((an*an-1+1)*an-2+1)*an-3+1.........)*a1
那剩下来就简单了。
mykeping 发布时间:
2010-5-1722:
39:
43
void main()
int max,min,z,i,a[50000],s;
scanf("%d",&z);
for(i=0;;i++)
scanf("%d",&a[i]);
if(a[i]==0)
{ s=i;
break;
max=a[0];
min=a[1];
for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:mykeping 发布时间:2010-5-1722:39:34#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1519:07:44#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1516:14:48以下是极差概念,指一组数据中最大数据与最小数据的差,在统计中常用极差来刻画一组数据的离散程度。以及表示,R=Xmax-Xmin。又称全距或范围误差。反映的是变量分布的变异范围和离散幅度,在总体中任何两个单位的标准值之差都不能超过极差。同时,它能体现一组数据波动的范围。 如 12 12 13 14 16 21 这组数的极差就是 21-12=9与题中不同,但是题中结果似乎是这么的来的,请版主提示评论人:leijf2010 发布时间:2010-5-1421:54:01我不知道自己的解法对不对,因为我还有点不明白这个题目的意思,如果是随便擦出其中的两个数,那么最后那个最大值和最小值是不是也要算上擦去两个数的乘积加1? 如:1 2 3,如果擦去的是1,2,那么加入的就是3。如果擦去2,3,加入的值就是7,那么最大值和最小值之差就是6。而不是2?自己的一些想法,不知道对不对?#include#includeusing namespace std;#define N 50000vectorb;int main(){ int A[N]; int n,j,k,l,m,sum; int max,min; int*p; cin>>n; for(int i=0;i { if(cin>>A[i]) b.push_back(A[i]); } k=b.size()-1; max=b[k]; min=b[k]; for(j=b.size()-1;j>0;j--) { l=b[j]; m=b[j-1]; if(m>max) max=m; else min=m; sum=l*m+1; p=&b[j]; b.erase(p-1,p); b.push_back(sum); cout< } cout< return 0;}评论人:hanhan0121 发布时间:2010-3-723:54:41#includevoid main(){ int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t; scanf("%d",&N); t=N; for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:mykeping 发布时间:2010-5-1722:39:34#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1519:07:44#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1516:14:48以下是极差概念,指一组数据中最大数据与最小数据的差,在统计中常用极差来刻画一组数据的离散程度。以及表示,R=Xmax-Xmin。又称全距或范围误差。反映的是变量分布的变异范围和离散幅度,在总体中任何两个单位的标准值之差都不能超过极差。同时,它能体现一组数据波动的范围。 如 12 12 13 14 16 21 这组数的极差就是 21-12=9与题中不同,但是题中结果似乎是这么的来的,请版主提示评论人:leijf2010 发布时间:2010-5-1421:54:01我不知道自己的解法对不对,因为我还有点不明白这个题目的意思,如果是随便擦出其中的两个数,那么最后那个最大值和最小值是不是也要算上擦去两个数的乘积加1? 如:1 2 3,如果擦去的是1,2,那么加入的就是3。如果擦去2,3,加入的值就是7,那么最大值和最小值之差就是6。而不是2?自己的一些想法,不知道对不对?#include#includeusing namespace std;#define N 50000vectorb;int main(){ int A[N]; int n,j,k,l,m,sum; int max,min; int*p; cin>>n; for(int i=0;i { if(cin>>A[i]) b.push_back(A[i]); } k=b.size()-1; max=b[k]; min=b[k]; for(j=b.size()-1;j>0;j--) { l=b[j]; m=b[j-1]; if(m>max) max=m; else min=m; sum=l*m+1; p=&b[j]; b.erase(p-1,p); b.push_back(sum); cout< } cout< return 0;}评论人:hanhan0121 发布时间:2010-3-723:54:41#includevoid main(){ int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t; scanf("%d",&N); t=N; for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
max=a[i];
if(min>a[i])
min=a[i];
printf("%d",max-min);
34
for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1519:07:44#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1516:14:48以下是极差概念,指一组数据中最大数据与最小数据的差,在统计中常用极差来刻画一组数据的离散程度。以及表示,R=Xmax-Xmin。又称全距或范围误差。反映的是变量分布的变异范围和离散幅度,在总体中任何两个单位的标准值之差都不能超过极差。同时,它能体现一组数据波动的范围。 如 12 12 13 14 16 21 这组数的极差就是 21-12=9与题中不同,但是题中结果似乎是这么的来的,请版主提示评论人:leijf2010 发布时间:2010-5-1421:54:01我不知道自己的解法对不对,因为我还有点不明白这个题目的意思,如果是随便擦出其中的两个数,那么最后那个最大值和最小值是不是也要算上擦去两个数的乘积加1? 如:1 2 3,如果擦去的是1,2,那么加入的就是3。如果擦去2,3,加入的值就是7,那么最大值和最小值之差就是6。而不是2?自己的一些想法,不知道对不对?#include#includeusing namespace std;#define N 50000vectorb;int main(){ int A[N]; int n,j,k,l,m,sum; int max,min; int*p; cin>>n; for(int i=0;i { if(cin>>A[i]) b.push_back(A[i]); } k=b.size()-1; max=b[k]; min=b[k]; for(j=b.size()-1;j>0;j--) { l=b[j]; m=b[j-1]; if(m>max) max=m; else min=m; sum=l*m+1; p=&b[j]; b.erase(p-1,p); b.push_back(sum); cout< } cout< return 0;}评论人:hanhan0121 发布时间:2010-3-723:54:41#includevoid main(){ int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t; scanf("%d",&N); t=N; for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1519:07:44#includevoid main(){ int max,min,z,i,a[50000],s; scanf("%d",&z); for(i=0;;i++) { scanf("%d",&a[i]); if(a[i]==0) { s=i; break; } } max=a[0]; min=a[1]; for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1516:14:48以下是极差概念,指一组数据中最大数据与最小数据的差,在统计中常用极差来刻画一组数据的离散程度。以及表示,R=Xmax-Xmin。又称全距或范围误差。反映的是变量分布的变异范围和离散幅度,在总体中任何两个单位的标准值之差都不能超过极差。同时,它能体现一组数据波动的范围。 如 12 12 13 14 16 21 这组数的极差就是 21-12=9与题中不同,但是题中结果似乎是这么的来的,请版主提示评论人:leijf2010 发布时间:2010-5-1421:54:01我不知道自己的解法对不对,因为我还有点不明白这个题目的意思,如果是随便擦出其中的两个数,那么最后那个最大值和最小值是不是也要算上擦去两个数的乘积加1? 如:1 2 3,如果擦去的是1,2,那么加入的就是3。如果擦去2,3,加入的值就是7,那么最大值和最小值之差就是6。而不是2?自己的一些想法,不知道对不对?#include#includeusing namespace std;#define N 50000vectorb;int main(){ int A[N]; int n,j,k,l,m,sum; int max,min; int*p; cin>>n; for(int i=0;i { if(cin>>A[i]) b.push_back(A[i]); } k=b.size()-1; max=b[k]; min=b[k]; for(j=b.size()-1;j>0;j--) { l=b[j]; m=b[j-1]; if(m>max) max=m; else min=m; sum=l*m+1; p=&b[j]; b.erase(p-1,p); b.push_back(sum); cout< } cout< return 0;}评论人:hanhan0121 发布时间:2010-3-723:54:41#includevoid main(){ int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t; scanf("%d",&N); t=N; for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
zhichen 发布时间:
2010-5-1519:
07:
44
for(i=0;i { if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1516:14:48以下是极差概念,指一组数据中最大数据与最小数据的差,在统计中常用极差来刻画一组数据的离散程度。以及表示,R=Xmax-Xmin。又称全距或范围误差。反映的是变量分布的变异范围和离散幅度,在总体中任何两个单位的标准值之差都不能超过极差。同时,它能体现一组数据波动的范围。 如 12 12 13 14 16 21 这组数的极差就是 21-12=9与题中不同,但是题中结果似乎是这么的来的,请版主提示评论人:leijf2010 发布时间:2010-5-1421:54:01我不知道自己的解法对不对,因为我还有点不明白这个题目的意思,如果是随便擦出其中的两个数,那么最后那个最大值和最小值是不是也要算上擦去两个数的乘积加1? 如:1 2 3,如果擦去的是1,2,那么加入的就是3。如果擦去2,3,加入的值就是7,那么最大值和最小值之差就是6。而不是2?自己的一些想法,不知道对不对?#include#includeusing namespace std;#define N 50000vectorb;int main(){ int A[N]; int n,j,k,l,m,sum; int max,min; int*p; cin>>n; for(int i=0;i { if(cin>>A[i]) b.push_back(A[i]); } k=b.size()-1; max=b[k]; min=b[k]; for(j=b.size()-1;j>0;j--) { l=b[j]; m=b[j-1]; if(m>max) max=m; else min=m; sum=l*m+1; p=&b[j]; b.erase(p-1,p); b.push_back(sum); cout< } cout< return 0;}评论人:hanhan0121 发布时间:2010-3-723:54:41#includevoid main(){ int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t; scanf("%d",&N); t=N; for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
if(max max=a[i]; if(min>a[i]) min=a[i]; } printf("%d",max-min); }评论人:zhichen 发布时间:2010-5-1516:14:48以下是极差概念,指一组数据中最大数据与最小数据的差,在统计中常用极差来刻画一组数据的离散程度。以及表示,R=Xmax-Xmin。又称全距或范围误差。反映的是变量分布的变异范围和离散幅度,在总体中任何两个单位的标准值之差都不能超过极差。同时,它能体现一组数据波动的范围。 如 12 12 13 14 16 21 这组数的极差就是 21-12=9与题中不同,但是题中结果似乎是这么的来的,请版主提示评论人:leijf2010 发布时间:2010-5-1421:54:01我不知道自己的解法对不对,因为我还有点不明白这个题目的意思,如果是随便擦出其中的两个数,那么最后那个最大值和最小值是不是也要算上擦去两个数的乘积加1? 如:1 2 3,如果擦去的是1,2,那么加入的就是3。如果擦去2,3,加入的值就是7,那么最大值和最小值之差就是6。而不是2?自己的一些想法,不知道对不对?#include#includeusing namespace std;#define N 50000vectorb;int main(){ int A[N]; int n,j,k,l,m,sum; int max,min; int*p; cin>>n; for(int i=0;i { if(cin>>A[i]) b.push_back(A[i]); } k=b.size()-1; max=b[k]; min=b[k]; for(j=b.size()-1;j>0;j--) { l=b[j]; m=b[j-1]; if(m>max) max=m; else min=m; sum=l*m+1; p=&b[j]; b.erase(p-1,p); b.push_back(sum); cout< } cout< return 0;}评论人:hanhan0121 发布时间:2010-3-723:54:41#includevoid main(){ int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t; scanf("%d",&N); t=N; for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
2010-5-1516:
48
以下是极差概念,
指一组数据中最大数据与最小数据的差,在统计中常用极差来刻画一组数据的离散程度。
以及表示,R=Xmax-Xmin。
又称全距或范围误差。
反映的是变量分布的变异范围和离散幅度,在总体中任何两个单位的标准值之差都不能超过极差。
同时,它能体现一组数据波动的范围。
如
12 12 13 14 16 21
这组数的极差就是
21-12=9
与题中不同,但是题中结果似乎是这么的来的,请版主提示
leijf2010 发布时间:
2010-5-1421:
54:
01
我不知道自己的解法对不对,因为我还有点不明白这个题目的意思,如果是随便擦出其中的两个数,那么最后那个最大值和最小值是不是也要算上擦去两个数的乘积加1?
如:
1 2 3,如果擦去的是1,2,那么加入的就是3。
如果擦去2,3,加入的值就是7,那么最大值和最小值之差就是6。
而不是2?
自己的一些想法,不知道对不对?
#define N 50000
vectorb;
int A[N];
int n,j,k,l,m,sum;
int*p;
cin>>n;
for(int i=0;i { if(cin>>A[i]) b.push_back(A[i]); } k=b.size()-1; max=b[k]; min=b[k]; for(j=b.size()-1;j>0;j--) { l=b[j]; m=b[j-1]; if(m>max) max=m; else min=m; sum=l*m+1; p=&b[j]; b.erase(p-1,p); b.push_back(sum); cout< } cout< return 0;}评论人:hanhan0121 发布时间:2010-3-723:54:41#includevoid main(){ int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t; scanf("%d",&N); t=N; for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
if(cin>>A[i])
b.push_back(A[i]);
k=b.size()-1;
max=b[k];
min=b[k];
for(j=b.size()-1;j>0;j--)
l=b[j];
m=b[j-1];
if(m>max)
max=m;
else
min=m;
sum=l*m+1;
p=&b[j];
b.erase(p-1,p);
b.push_back(sum);
cout< } cout< return 0;}评论人:hanhan0121 发布时间:2010-3-723:54:41#includevoid main(){ int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t; scanf("%d",&N); t=N; for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
cout< return 0;}评论人:hanhan0121 发布时间:2010-3-723:54:41#includevoid main(){ int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t; scanf("%d",&N); t=N; for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
hanhan0121 发布时间:
2010-3-723:
41
int m,N,a[5000],b[5000],i,j,p,q,max,min,y,t;
scanf("%d",&N);
t=N;
for(i=0;i scanf("%d",&a[i]); for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
for(i=0;i b[i]=a[i]; while(N>0) { m=a[0]; q=0; for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
while(N>0)
m=a[0]; q=0;
for(i=1;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; m=a[0]; q=0; for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
if(a[i]>m)
m=a[i];
q=i;
p=a[N-1]; a[N-1]=a[q]; a[q]=p;
for(i=0;i if(a[i]>m) { m=a[i]; q=i; } p=a[N-1]; a[N-1]=a[q]; a[q]=p; a[N-2]=a[N-2]*a[N-1]+1;N=N-1; } min=a[0]; N=t; while(N>0) { m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
a[N-2]=a[N-2]*a[N-1]+1;N=N-1;
min=a[0];
N=t;
m=b[0]; q=0;
for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; m=b[0]; q=0; for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
m=b[i];
p=b[N-1]; b[N-1]=b[q]; b[q]=p;
for(i=0;i if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
if(b[i] { m=b[i]; q=i; } p=b[N-1]; b[N-1]=b[q]; b[q]=p; b[N-2]=b[N-2]*b[N-1]+1; N=N-1; } max=b[0]; printf("%d\n",(max-min));}评论人:kkskssk 发布时间:2008-11-2514:09:16#include #define M 20using namespace std;int Max(int a[],int n) //求最大值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
b[N-2]=b[N-2]*b[N-1]+1;
N=N-1;
max=b[0];
printf("%d\n",(max-min));
kkskssk 发布时间:
2008-11-2514:
16
#define M 20
int Max(int a[],int n) //求最大值
if (n==1)
return a[0];
for(int j=0;j { for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
for(int i=0;i { if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
if (a[i] { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Max(a,n-1); }}int Min(int a[],int n) //求最小值{ if (n==1) { return a[0]; } else { for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
int p=a[i];
a[i]=a[i+1];
a[i+1]=p;
a[n-2]=a[n-2]*a[n-1]+1;
return Max(a,n-1);
int Min(int a[],int n) //求最小值
for(int j=0;j { for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
for(int i=0;i { if (a[i]>a[i+1]) { int p=a[i]; a[i]=a[i+1]; a[i+1]=p; } } } a[n-2]=a[n-2]*a[n-1]+1; return Min(a,n-1); }}int main (){ int n,max,min,lim,a[M],b[M]; cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
if (a[i]>a[i+1])
return Min(a,n-1);
int main ()
int n,max,min,lim,a[M],b[M];
cout<<"请输入元素个数(n<="<"; cin>>n; for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
";
for (int i=0;i { cin>>a[i]; b[i]=a[i]; } max=Max(a,n); min=Min(b,n); lim=max-min; cout<<"数据极值为:"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
max=Max(a,n);
min=Min(b,n);
lim=max-min;
cout<<"数据极值为:
"< return 0; }评论人:神愚嘻嘻 发布时间:2008-10-2421:49:33program ex1;var x,y,n,m,l,i,j,k,t,zong:longint; a,b:array[1..100,1..50000] of integer; min,max:longint; shu:array[1..100] of longint;procedure sort(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=a[k,(l+r) shr 1]; repeat while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
神愚嘻嘻 发布时间:
2008-10-2421:
49:
33
program ex1;
var
x,y,n,m,l,i,j,k,t,zong:
longint;
a,b:
array[1..100,1..50000] of integer;
min,max:
shu:
array[1..100] of longint;
procedure sort(l,r:
longint);
ii,jj,y,x:
begin
ii:
=l; jj:
=r;
x:
=a[k,(l+r) shr 1];
repeat
while a[k,ii]=ii+1; while a[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=a[k,ii];a[k,ii]:=a[k,jj];a[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
=ii+1;
while a[k,jj]>x do jj:
=jj-1;
if ii<=jj then
y:
=a[k,ii];a[k,ii]:
=a[k,jj];a[k,jj]:
=y;
=ii+1; jj:
end;
until ii>jj;
if l if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
if iiend;procedure sort1(l,r:longint);var ii,jj,y,x:longint;begin ii:=l; jj:=r; x:=b[k,(l+r) shr 1]; repeat while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
procedure sort1(l,r:
=b[k,(l+r) shr 1];
while b[k,ii]=ii+1; while b[k,jj]>x do jj:=jj-1; if ii<=jj then begin y:=b[k,ii];b[k,ii]:=b[k,jj];b[k,jj]:=y; ii:=ii+1; jj:=jj-1; end; until ii>jj; if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
while b[k,jj]>x do jj:
=b[k,ii];b[k,ii]:
=b[k,jj];b[k,jj]:
if l if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
if iiend;{============main===========}begin assign(input,'jj.in');reset(input); zong:=0; repeat zong:=zong+1; readln(shu[zong]); for i:=1 to shu[zong] do readln(a[zong,i]); for i:=1 to shu[zong] do b[zong,i]:=a[zong,i]; until shu[zong]=0; assign(output,'jj.out');rewrite(output); zong:=zong-1; for k:=1 to zong do begin min:=0; max:=0; sort(1,shu[k]); sort1(1,shu[k]); for j:=1 to shu[k]-1 do begin a[k,j+1]:=a[k,j+1]*a[k,j]+1; sort(j+1,shu[k]); end; max:=a[k,shu[k]]; for j:=shu[k] downto 2 do begin b[k,j-1]:=b[k,j-1]*b[k,j]+1; sort1(1,j-1); end; min:=b[k,1]; writeln(max-min); end; close(input);close(output);end.
{============main===========}
assign(input,'jj.in');reset(input);
zong:
=0;
=zong+1;
readln(shu[zong]);
for i:
=1 to shu[zong] do readln(a[zong,i]);
=1 to shu[zong] do b[zong,i]:
=a[zong,i];
until shu[zong]=0;
assign(output,'jj.out');rewrite(output);
=zong-1;
for k:
=1 to zong do
min:
=0; max:
sort(1,shu[k]); sort1(1,shu[k]);
for j:
=1 to shu[k]-1 do
a[k,j+1]:
=a[k,j+1]*a[k,j]+1;
sort(j+1,shu[k]);
max:
=a[k,shu[k]];
=shu[k] downto 2 do
b[k,j-1]:
=b[k,j-1]*b[k,j]+1;
sort1(1,j-1);
=b[k,1];
writeln(max-min);
close(input);close(output);
end.
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