# A hard puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25563    Accepted Submission(s): 9084

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

```#include
int main()
{
long a,b;
int temp;
while(scanf("%ld%ld",&a,&b)!=EOF)
{
a=a%10;
b=(b>4)?b%4:b;
if(b==0)
{
b=4;
}
for(temp=1;b>0;b--)
{
temp=temp*a;
}
printf("%dn",temp%10);
}
return 0;
}
```

一道找规律的题目，研究一下会发现末位数都是有规律的，都是4的倍数

1是1、1、1、1
2是2、4、8、6
3是3、9、7、1
4是4、6、4、6
5是5、5、5、5
6是6、6、6、6
7是7、9、3、1
8是8、4、2、6
9是9、1、9、1