Rightmost Digit-优快云博客

本文介绍了一种快速求解正整数N的N次方最右位数字的方法，并通过观察发现了该运算结果的周期性规律。利用这一规律，可以通过简单的算法高效地计算出任意大整数N的N次方的最右位数字。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Sample Output

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.

可以发现一个规律：

当 n = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 27 28 29 30 31 ...

rdigit = 1 4 7 6 5 6 3 6 9 0 1 6 3 6 5 6 7 4 9 0 1 4 7 6 5 6 3 6 9 0 ...

所以是以20为周期的规律。

得到这个规律，算法就变得很easy

#include<iostream>
using namespace std;
int main()
{
    int  rdigit[20] = {0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9};
    int n;
    cin>>n;
    int res[n];
    for(int i=0;i<n;i++)
    {
        int temp;
        cin>>temp;
        res[i]=rdigit[temp%20];
    }    
    for(int i=0;i<n;i++)
    {
        cout<<res[i]<<endl;
    }
    system("pause");
    return 0;
}