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Rightmost Digit

时间:2018-10-04 13:22:04      阅读:169      评论:0      收藏:0      [点我收藏+]

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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:
 
技术分享图片
 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

标签:targe   pre   each   eve   返回   图片   otto   计算   style   

原文地址:https://www.cnblogs.com/Luckykid/p/9742011.html

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