Rightmost Digit

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.

 

代码:

 

#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;
}