Rightmost Digit（最后一位数字）

Description

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. 

题目意思：
求n的n次方的最后一位数字

 #include<stdio.h>
 #include<string.h>
 int main()
 {
     long long int n,i,j,ans,t,sum;;
     scanf("%lld",&t);
     while(t--)
     {
         scanf("%lld",&n);
         sum=;
         ans=;
         n=n%;//找出规律,20个数一次循环
         if(n==)
         {
             n=;
         }//将范围缩小到20以内
         for(i=;i<n;i++)
         {
             sum=ans*n;
             ans=sum%;//(只要最后的一位)
         }
         ans=ans%;
         printf("%lld\n",ans);
     }
     return ;
 }