hdu1061 Rightmost Digit(找规律+快速幂)_替换数字数字两边需要固定找规律 hdu-优快云博客

本文介绍了求解正整数N的N次方最后一位数字的两种方法：一是通过观察规律直接映射查找，二是利用快速幂算法结合模运算。通过两段C++代码示例，展示了如何高效解决这一数学问题。

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题目：
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.

题意：找到n^n的最后一个数字
方法一：找规律 ans[20] = {0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9};
方法二：用快速幂的方法，结果为快速幂之后对10取余的数

#include
using namespace std;

int main(int argc, const char * argv[]) {
int ans[20] = {0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9};
int n,t;
while(cin >> t){
while (t–) {
cin >> n;
cout << ans[n%20] <<endl;
}
}
return 0;
}
`
2

//
//  main.cpp
//  hdu1061
//
//  Created by zhan_even on 2018/10/28.
//  Copyright © 2018年 zhan_even. All rights reserved.
//

#include <iostream>
using namespace std;
typedef long long ll;
ll powerMod(ll a, ll n, int m){
    ll res = 1;
    while (n) {
        if (n&1) {
            res *= a;
            res %=m;
        }
        a *= a;
        a %= m;
        n >>= 1;
    }
    return res;
}

int main(){
    int t,n;
    cin>>t;
    while(t--) {
        cin >> n;
        
        cout << powerMod(n, n, 10) << endl;
    }
    return 0;
}