N^N次方最后一位数字的求法

Problem 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

#include<iostream>using namespace std;int main(){int a[10][2] ={{0},{1},{4,6},{7,3},{6},{5},{6},{3,7},{6,4},{9}};//每个不同数字的N次方的结果的可能性列举int rows,tmp;cin>>rows;int *arr = new int[rows];for(int i = 0;i<rows;i++)cin>>arr[i];for(int i = 0;i<rows;i++){tmp = arr[i]%10;switch(tmp){case 0:cout<<a[0][0]<<endl;break;case 1:cout<<a[1][0]<<endl;break;case 2:{if((arr[i]-tmp)%20 == 0)//减去结尾数字求得除以20的余数是否为0cout<<a[2][0]<<endl;elsecout<<a[2][1]<<endl;break;}case 3:{if((arr[i]-tmp)%20 == 0)cout<<a[3][0]<<endl;elsecout<<a[3][1]<<endl;break;}case 4:cout<<a[4][0]<<endl;break;case 5:cout<<a[5][0]<<endl;break;case 6:cout<<a[6][0]<<endl;break;case 7:{if((arr[i]-tmp)%20 == 0)cout<<a[7][0]<<endl;elsecout<<a[7][1]<<endl;break;}case 8:{if((arr[i]-tmp)%20 == 0)cout<<a[8][0]<<endl;elsecout<<a[8][1]<<endl;break;}case 9:cout<<a[9][0]<<endl;break;}}return 0;}