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:
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 }
View Code
Rightmost Digit
原文：https://www.cnblogs.com/Luckykid/p/9742011.html
