西工大C语言POJ习题答案.docx
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一.1.第一季10题全(注:
第五题问题已经解决,确认AC!
)
#include
int main()
{
int a,b,sum;
scanf("%d%d",&a,&b);
sum=a+b;
printf("%d\n",sum);
return 0;
}
2.
#include
#define PI 3.1415926
int main(){
double r,h,l,s,sq,vq,vz;
scanf("%lf%lf",&r,&h);
l=2*PI*r;
s=PI*r*r;
sq=4*PI*r*r;
vq=4*PI*r*r*r/3;
vz=s*h;
printf("%.2lf\n%.2lf\n%.2lf\n%.2lf\n%.2lf\n",l,s,sq,vq,vz);
return 0;
}
3.
#include
int main(){
int a,b,c;
double d,e;
scanf("%d%d%d",&a,&b,&c);
d=a+b+c;
e=d/3;
printf("%lf\n%lf\n",d,e);
return 0;
}
4.
#include
int main(){
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
if(a
a=b;
if(a a=c;
printf("%d\n",a);
return 0;
}
5.
#include
int main()
{
int i=0,j=0,k=1;
char a[6];
while((a[i]=getchar())!
='\n')
{
i++;
}
for(;i>0;i--){
if(a[j]==a[i-1]){
j++;
continue;}
else {k=0;
break;}
}
if(k==1)
printf("yes\n");
else
printf("no\n");
}
6.
#include
int main()
{
double a,c;
scanf("%lf",&a);
switch((int)a/10)
{
case 0:
c=a*0.1;break;
case 1:
c=(a-10)*0.075+10*0.1;break;
case 2:
case 3:
c=(a-20)*0.05+10*0.075+10*0.1;break;
case 4:
case 5:
c=(a-40)*0.03+20*0.05+10*0.075+10*0.1;break;
case 6:
case 7:
case 8:
case 9:
c=(a-60)*0.015+20*0.03+20*0.05+10*0.075+10*0.1;break;
default:
c=(a-100)*0.01+40*0.015+20*0.03+20*0.05+10*0.075+10*0.1;
}
printf("%lf\n",c);
return 0;
}
7.
#include
int main()
{
double a,b,c;
scanf("%lf",&a);
c=(int)a;
if(a>c)a=c+1;
if(a>15)
b=(a-15)*2.1+7+13*1.5;
else {
if(a>2)
b=(a-2)*1.5+7;
else b=7;
}
printf("%lf\n",b);
return 0;
}
8.
#include
int main()
{
int a,b,c,e,f=30,g=31,n;
scanf("%d-%d-%d",&a,&b,&c);
if((a%400==0)||(a%100!
=0&&a%4==0))
e=29;
else
e=28;
switch (b)
{
case 1:
n=c;break;
case 2:
n=g+c;break;
case 3:
n=g+e+c;break;
case 4:
n=g+e+g+c;break;
case 5:
n=g+e+g+f+c;break;
case 6:
n=g+e+g+f+g+c;break;
case 7:
n=g+e+g+f+g+f+c;break;
case 8:
n=g+e+g+f+g+f+g+c;break;
case 9:
n=g+e+g+f+g+f+g+g+c;break;
case 10:
n=g+e+g+f+g+f+g+g+f+c;break;
case 11:
n=g+e+g+f+g+f+g+g+f+g+c;break;
default:
n=g+e+g+f+g+f+g+g+f+g+f+c;
}
printf("%d\n",n);
return 0;
}
9.
#include
int main()
{
int x;
scanf("%d",&x);
if(x>=90&&x<=100)
printf("A\n");
else if (x>=80)
printf("B\n");
else if (x>=70)
printf("C\n");
else if (x>=60)
printf("D\n");
else
printf("E\n");
return 0;
}
10.
#include
int main()
{
double x,y,s;
scanf("%lf,%lf",&x,&y);
s=(x+2)*(x+2)+(y-2)*(y-2);
if(s>1){
s=(x+2)*(x+2)+(y+2)*(y+2);
if(s>1){
s=(x-2)*(x-2)+(y+2)*(y+2);
if(s>1){
s=(x-2)*(x-2)+(y-2)*(y-2);
if(s>1){printf("0\n");return 1;}
}
}
}
printf("10\n");
return 0;
}
二。
第二季15题全
11.
#include
intmain()
{
intx=0,a=0,c=0,i=1;
scanf("%d%d",&x,&a);
c=x;
for(i=1;i
c=(c*x)%1000;
};
if(c<100){
if(c<10)
printf("00%d\n",c);
else
printf("0%d\n",c);
};
if(c>=100)
printf("%d\n",c);
return0;
}
12.
#include
intmain()
{
inti=1,a=0,b=0,c=1016,k=1;
for(;k=1;){
c=c+4;
a=c;
b=c;
for(i=1;i<=4;i++){
a=a/4*5+1;
if(a%4!
=0){
k=0;
;};
};
a=a/4*5+1;
if(k==1){
printf("%d%d\n",a,b);
break;
};
};
return0;
}
13。
。
#include
#include
intmain()
{
inti=1,n=0,b=0;
doublej=0,k=0,m=0;
scanf("%d",&n);
for(;i<=n;i++){
b=i%2-1;
k=pow(-1,b);
m=pow(2,i);
j+=k*m/(m-k)/(2*m+k);
};
printf("%.6lf\n",j);
return0;
}
14.
#include
#include
intmain()
{
inta=0,b=0,i=0,t;
scanf("%d%d",&a,&b);
if(a>b){t=a;a=b;b=t;}
for(;a<=b;a++)
{intm=1;
for(i=2;i<=sqrt(a);i++){
if(a%i==0){
m=0;
break;};
}
if(m==1)
printf("%d",a);
}
printf("\n");
return0;
}
15.
#include
#include
intmain()
{
inta=0,b=0,c=0;
scanf("%d",&a);
b=(int)pow(a,2);
c=(int)pow(a,3);
if(a%2)
{printf("%d*%d*%d=%d=%d",a,a,a,c,b-a/2*2);
intn=1;
for(;n printf("+%d",b-a/2*2+2*n);
}
printf("\n");
};
if(a%2==0)
{
intn=1;
printf("%d*%d*%d=%d=%d",a,a,a,c,b+1-a);
for(;n printf("+%d",b+1-a+n*2);
}
printf("\n");
};
return0;
}
16
#include
#include
int