1、); elseNOn习题1-7 #include double a; a=n*95.0; if(a0) count+; n=n/10;,count);习题2-2 for(int i=100;i=999;i+) a=i%10; b=i/10%10; c=i/100; if(i=a*a*a+b*b*b+c*c*c),i);习题2-3#include=0;i-) for(int j=n-i;j0;j-) for(int j=2*i-1;#n习题2-5文件题,南邮竞赛基本不涉及。习题2-6 int i,n; double sum=1.0; for(i=2;=n; sum+=(1.0/i);,sum);
2、习题2-7 int t=-1; double a=1.0,sum=1.0; while(fabs(a)=0.000001) a=1.0/(a+2); a=a*t; sum=sum+a; t=t*(-1);%.9lfn习题2-8 int n,m,temp,kase=0;%d%dn,&m) double s=0; if(nm)temp=n;n=m;m=temp; if(n=0&m=0) break; for(int i=n;=m; s+=1.0/i/i;%.5fn,+kase,s);习题2-9 printf的特殊用法:对于m.n的格式可以用如下方法表示 char ch20;%*.*sn,m,n,c
3、h); 前边的*定义的是总的宽度,后边的定义的是输出的个数。分别对应外面的参数m和n 。 这种方法的好处是可以在语句之外对参数m和n赋值,从而控制输出格式。 %.*lfn,c,(double)a/b);习题2-10#define FOR(i) for(i=1;10; int a,b,c,d,e,f,g,h,i; FOR(a) FOR(b) FOR(c) FOR(d) FOR(e) FOR(f) FOR(g) FOR(h) FOR(i) if(2*(a*100+b*10+c)=(d*100+e*10+f)&3*(a*100+b*10+c)=(g*100+h*10+i) if(a!=b)&(a!=
4、c)&=d)&=e)&=f)&=g)&=h)&=i)& (b!(b! (c!(c! (d!(d!(e! (e!(f!(g!(h!=i)%dn%dn%dn,a*100+b*10+c,d*100+e*10+f,g*100+h*10+i); 第三章习题3-1string.h int num80; char str81; int t;t); while(t-) int sum=0;%s,str); str0=O?num0=1:num0=0; for(int i=1;strlen(str); stri=numi=numi-1+1:numi=0; sum+=numi;,sum+num0);习题3-2cty
5、pe.hdouble M(char x) if(x=C) return 12.01;H return 1.008; return 16.00;N return 14.01; char str100; int N,i,j; double sum;N); while(N-) sum=0; for(i=0; if(isalpha(stri) if(stri+11 if(stri+2 sum=sum+M(stri)*(stri+1-0)*10+stri+2- i=i+2; sum=sum+M(stri)*(stri+1- i+; sum+=M(stri);习题3-3 char s1000000; in
6、t a10000;,s) int b10=; for (int i=0;10000;i+) bsi-+;9;i+) printf(%d , bi);, b9);习题3-4char s85; while(scanf( int len=strlen(s);=len;+i) if(len%i=0) int k; for(k=1;k+k) if(sk!=sk%i) break; if(k=len) printf(break;习题3-5 #include const int LEN=5; const int MAX=100; const int y=0,0,1,-1; const int x=-1,1,
7、0,0; char mapLENLEN; int tra110; bool legal(int pos) return 0=pos&posLEN; void Pmap() for(int cow=0;cowcow+)%c,mapcow0); for(int col=1;colcol+) %c,mapcowcol); int main() traA=0;B=1;R=2;L=3; bool first=true; int Case=0; int bx,by; while(gets(map0) if(map00=Z)break; gets(mapcol); for(int i=0; for(int
8、j=0;jj+) if(mapij= ) bx=i;by=j; bool ok=true; char c; while(scanf(c),c! if(!ok)continue; int nx=bx+xtrac,ny=by+ytrac;legal(nx)|!legal(ny) ok=false; continue; mapbxby=mapnxny; mapnxny=; bx=nx;by=ny; getchar(); if(first) first=false;Puzzle #%d:,+Case); if(ok) Pmap();This puzzle has no final configurat
9、ion./n习题3-6int first=1;char map1212;struct point int x,y; int r,c; str111;r,&c),r,c)r;,mapi); int num=0; for(int j=0;c; if(mapij!* if(mapij-1=|j-1 strnum.y=j; strnum.x=i; strnum.r=1; num+; else strnum.r=0; if(mapi-1j=|i-1 strnum.y=j; strnum.c=1; else strnum.c=0; first=0;Across:num; if(stri.r) for(in
10、t j=stri.y; if(mapstri.xj=,mapstri.xj);Down: if(stri.c) for(int j=stri.x; if(mapjstri.y=,mapjstri.y);习题3-7const int N = 1005;const int M = 105;int n, m, cntM;char dnaMN;void init() , &m, & for (int i = 0; i m; i+) scanf(, dnai);void solve() int ans = 0; else printf( n; i+) int Max = 0, id; memset(cn
11、t, 0, sizeof(cnt); for (int j = 0; j Max) Max = cnttmp; id = tmp; else if (cnttmp = Max & tmp cstringusing namespace std; const int N=3005; int aN,vN; int main() int n, m, cnt;m) cnt = 0; memset(v,0,sizeof(v);%d/%d = %d.,n,m,n/m); n=n%m; while(!vn) a+cnt=(n*10)/m; vn=cnt; n = n * 10 % m;for(int i=1;
12、=cnt&51; +i) if(i=vn) printf(, ai); if(i=50) printf(.)n %d = number of digits in repeating cyclenn, cnt- vn+1);习题3-9iostream#define SIZE 1100000char subSIZE, baseSIZE;%s%s,base,sub) != EOF) int len = strlen(sub); int base_len = strlen(base); int j = 0; if(len base_len)Non i=len) break; if(j = len) printf(Yesn习题3-10algorithmstruct node int h,w; bool operator (const node& r) const ret
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