Rightmost Digit 快速幂

Given a positive integer N, you should output the most right digit of N^N.
Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).
Output
For each test case, you should output the rightmost digit of N^N.
Sample Input
2
3
4
Sample Output
7
6

Hint
In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7.
In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.

一个类似于快速幂的题目，求N的N次方的个位数，相当于只对个位数求高次幂，利用快速幂的思想，可以进行化简，不然会超时。
涉及到快速幂的思想，就是利用幂的关系进行化简，当幂为偶数时可以把底数化为平方然后将幂除以二，如果幂为奇数就提出一个底数然后再当作偶数进行化简，直到幂为0时结束。
AC代码：

#include<stdio.h> 
long long digital(long long a,long long b) 
{     
        long long ans=1;     
        while(b)    
       {         
           if(b%2==1)             
          ans=(ans*a)%10;         
          a=(a*a)%10;         
          b/=2;     
      }     
   return ans; 
} 
int main() 
{     
     long long n,m;     
     scanf("%I64d",&n);     
     while(n--)     
    {         
          scanf("%I64d",&m);         
          printf("%I64d\n",digital(m,m));     
    }     
  return 0; 
}