Given a positive integer N, you should output the most right digit of N^N.       

        

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).       

              

For each test case, you should output the rightmost digit of N^N.       

              

              

 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. 

方法一：

 #include <iostream>

 using namespace std;

 int a[]={,,,,,,,,,,,,,,,,,,,,};

 int main()

 {

     int b,n;

     cin>>b;

     while(b--)

     {

        cin>>n;

        cout<<a[n%]<<endl;

     }

     return  ;

 }

 #include<stdio.h>

 int my_power(int m, int n); // 求m的n次方的尾数

 int main()

 {

     int cases, n;

     scanf("%d", &cases);

     while(cases--)

     {

         scanf("%d", &n);

         printf("%d\n", my_power(n, n));

     }

     return ;

 }

 int my_power(int m, int n)

 {

     m = m%;

     if(n == )

         return m;

     if(n% == )

         return ( my_power(m*m, n/) ) % ;

     else

         return ( my_power(m*m, n/)*m ) % ;

 }