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=1;
        ans=1;
        n=n%20;//找出规律,20个数一次循环
        if(n==0)
        {
            n=20;
        }//将范围缩小到20以内
        for(i=0;i<n;i++)
        {
            sum=ans*n;
            ans=sum%10;//(只要最后的一位)
        }
        ans=ans%10;
        printf("%lld\n",ans);
    }
    return 0;
}