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