HDOJ 1061 Rightmost Digit 13.04.21周赛结题报告

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原创最新推荐文章于 2026-11-08 21:05:35 发布 903 阅读 AI 写同款· GEO 优化 ›

5个4000比特大状态置换算法的设计与分析下面是 5 个概念性、实验性的 4000 位可逆置换设计。它们都保证是双射，但未经过密码分析，不可直接用于生产系统。风格分别覆盖：三维 SPN、Feistel、可逆元胞自动机、ARX、宽轨迹/Keccak 式结构。这 5 个设计分别代表了 SPN、Feistel、元胞自动机、ARX、宽轨迹 Keccak 五种不同风格，覆盖了主流大状态置换的构造思路。再次强调：它们只是实验性概念，若要用于实际协议，必须经过严格安全性分析。阅读详情

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23861 Accepted Submission(s): 9096

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：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int main ()
{
    int n,t,m,i,s;
    cin>>t;
    while(t--)
    {
        scanf("%d",&n);
        m=n%10;
        n=n%4;
        s=m*m*m*m;
        s=s%10;
        for (i=0; i<n; i++)
            s*=m;
        s=s%10;
        cout<<s<<endl;
    }
    return 0;
}
/*
1: 1                4: 6,4,6        7: 9,3,1,7,9
2: 4,8,6,2,4        5: 5            8: 4,2,6,8,4
3: 9,7,1,3,9        6: 6,           9: 1,9,1
*/