HDOJ--1061--Rightmost Digit

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40004    Accepted Submission(s): 15104

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

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.
 

 题意：给你一个数 求它的N次幂，就是提示上面说的那样。

思路：还是一道快速幂模板题，不过有一点需要认识到的就是N太大的时候要先对N进行取余，不然答案有可能会出现溢出错误。

ac代码：

#include<stdio.h>int  fun(int a){int ans=1,b=a;a=a%10;while(b){if(b%2)ans=(ans*a)%10;a=(a*a)%10;b/=2;}return ans;}int main(){int a,T;scanf("%d",&T);while(T--){scanf("%d",&a);printf("%d\n",fun(a));}return 0;}