Rightmost Digit(快速幂+同余定理)

题目链接
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
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.

#include<iostream>using namespace std;int f(int a,int b){    int ans=1;    a = a % 10;    while(b > 0){        if(b & 1)            /**                1.b & 1取b二进制的最低位，判断和1是否相同，相同返回1，否则返回0，可用于判断奇偶                2.b>>1//把b的二进制右移一位，即去掉其二进制位的最低位            */            ans = (ans * a) % 10;            b = b >> 1;            a = (a * a)%10;    }    return ans;}int main(){        int n,t;        int result;        cin>>t;        while(t--){            cin>>n;            result=f(n,n);//计算n的b次方            cout<<result<<endl;        }        return 0;}

快速幂+同余定理详细解析