Rightmost Digit（快速幂）

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. 

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  ;
 }

方法二：

 #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 ) % ;
 }

快速幂的时间复杂度是O(logn)，n = 10亿时，大约32次递归调用就能出结果，效率极大的提高了