问的问题如果有点低级请不要鄙视。。
我是新手。。
嗯下面的程序两个printf都不是我要的答案，第二个还输不出来，哪里错了
求指点谢谢

#include"stdio.h"
#include"stdlib.h"
#include"math.h"

int main()
{
int x,y,a,b,c,d,t;
while(scanf("%d%d",&x,&y)!=EOF)
{
a=pow(x,1)%10;
b=pow(x,2)%10;
c=pow(x,3)%10;
d=pow(x,4)%10;
t=y%4;
printf("%d%d%d%d",a,b,c,d);
switch(t)
 {
 case'1':
 printf("%d",a);
 break;
 case'2':
 printf("%d",b);
 break;
case'3':
 printf("%d",c);
  break;
 case'0':
 printf("%d",d);
  break;
 }
}
return 0;
}

谢谢，改动后可以运行了，，
但是提交的时候出现了问题，

这是题目的网址
http://acm.hdu.

我修改后的程序这样
#include"stdio.h"
#include"stdlib.h"
#include"math.h"

int main()
{
int x,y,a,b,c,d,t;
while(scanf("%d%d",&x,&y)!=EOF)
{
a=(int)pow(x,1)%10;
b=(int)pow(x,2)%10;
c=(int)pow(x,3)%10;
d=(int)pow(x,4)%10;
t=y%4;

switch(t)
 {
 case 1:
 printf("%d\n",a);
 break;
 case 2:
 printf("%d\n",b);
 break;
case 3:
 printf("%d\n",c);
  break;
 case 0:
 printf("%d\n",d);
  break;
 }
}
return 0;
}

提交不了怎么办。。。

Problem Description
lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.
this puzzle describes that: gave a and b,how to know the a^b's the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise.

 

Input
There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30)

 

Output
For each test case, you should output the a^b's last digit number.

 

Sample Input
7 66
8 800
 

Sample Output
9
6