HDU 1061 Rightmost Digit

最新推荐文章于 2019-04-18 20:08:35 发布

原创最新推荐文章于 2019-04-18 20:08:35 发布 · 301 阅读

0 ·

CC 4.0 BY-SA版权

ACM解题报告专栏收录该内容

24 篇文章

订阅专栏

本文探讨了如何通过寻找特定规律来快速计算任意正整数N的N次方的个位数，利用循环特性简化计算过程。

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.

输入3，输出3*3*3的个位；

输入4，输出4*4*4*4的个位；

可以找到规律，20个数一个循环。其循环为：1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0

代码如下：

#include <iostream>
using namespace std;

int s[20]={1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0};
int main(){
	int n,x,ans;
	cin >> n;
	while(n--){
		cin >> x;
		ans = s[x%20-1];
		cout << ans << endl;
	}
	return 0;
}