hdu---1061 Rightmost Digit

求解N^N最右位数

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最新推荐文章于 2024-03-23 19:11:30 发布

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分类专栏： ====ACM算法学习==== 找规律数学-基础

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本文介绍了一个算法问题：给定正整数N，输出N^N的最右一位数字。通过寻找循环节的方法来解决大整数运算过程中可能遇到的溢出问题，并提供了完整的C++代码实现。

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35487 Accepted Submission(s): 13456

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

简单的找循环节

注意中间会爆int！！！

#include <iostream>
#include <algorithm>
#include <string.h>
#include <math.h>
#include <stdio.h>
#include <queue>
using namespace std;
#define N 10010
#define Mod 10
#define LL long long
/*

 */
int f[100010];
int main(){
	//freopen("in.in","r",stdin);
	//freopen("out.out","w",stdout);
	int T;
	while(scanf("%d",&T)!=EOF){
		while(T--)
		{
			int n;scanf("%d",&n);
			//找循环节
			int t = n%Mod;
			int num=1;
			LL a = n;
			f[1] = t;
			while(1){
				a = (a*n)%Mod;
				num++;
				f[num] = a;
				if(a == t) break;
			}
			num--; //循环节为num
			//循环节后的余数
			f[0] = f[num];
			printf("%d\n",f[n%num]);
		}
	}
	return 0;
}