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).
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
234
Sample Output
76
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:
1 #include<iostream> 2 using namespace std; 3 int f(int a,int b){ 4 ????int ans=1; 5 ????a = a % 10; 6 ????while(b > 0){ 7 ????????if(b & 1) 8 ????????????/** 9 ????????????????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10 ????????????????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11 ????????????*/12 ????????????ans = (ans * a) % 10;13 ????????????b = b >> 1;14 ????????????a = (a * a)%10;15 ????}16 ????return ans;17 }18 int main(){19 ????????int n,t;20 ????????int result;21 ????????cin>>t;22 ????????while(t--){23 ????????????cin>>n;24 ????????????result=f(n,n);//计算n的b次方25 ????????????cout<<result<<endl;26 ????????}27 ????????return 0;28 }Rightmost Digit
原文地址：https://www.cnblogs.com/Luckykid/p/9742011.html