西工大C语言POJ习题答案

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;br eak;

case 4:

case 5:c=(a-40)*0.03+20*0.05+10*0.075+1 0*0.1;break;

case 6:

case 7:

case 8:

case 9:c=(a-60)*0.015+20*0.03+20*0.05+1 0*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

int main()

{

int x=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);

return 0;

}

12.

#include

int main()

{

int i=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;

};

};

return 0;

}

13。。

#include

#include

int main()

{