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[25]={0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0};
int main()
{
    int b,n; 
    cin>>b;
    while(b--)
    {
       cin>>n;
       cout<<a[n%20]<<endl; 
    }
    return 0 ;
}

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
", my_power(n, n));
    }
    
    return 0;
}

int my_power(int m, int n)
{
    m = m%10;
    if(n == 1)
        return m;
    if(n%2 == 0)
        return ( my_power(m*m, n/2) ) % 10;
    else
        return ( my_power(m*m, n/2)*m ) % 10;
}

    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