(hdu step 2.2.2)Joseph(约瑟夫环变形问题)

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本文介绍了一个经典的Joseph问题变种，即找到最小的步长m，确保所有坏人在第一个好人被处决前全部被淘汰。通过使用暴力枚举的方法，文章提供了一段C++代码实现，并解释了如何避免超时错误。

题目:

Joseph

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2078 Accepted Submission(s): 1204

Problem Description

The Joseph\\\\\\\'s problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n, standing in circle every mth is going to be executed and only the life of the last remaining person will be saved. Joseph was smart enough to choose the position of the last remaining person, thus saving his life to give us the message about the incident. For example when n = 6 and m = 5 then the people will be executed in the order 5, 4, 6, 2, 3 and 1 will be saved.

Suppose that there are k good guys and k bad guys. In the circle the first k are good guys and the last k bad guys. You have to determine such minimal m that all the bad guys will be executed before the first good guy.

Input

The input file consists of separate lines containing k. The last line in the input file contains 0. You can suppose that 0 < k < 14.

Output

The output file will consist of separate lines containing m corresponding to k in the input file.

Sample Input

3
4
0

Sample Output

5
30

Source

ACM暑期集训队练习赛（5）

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题目大意：

有2*k个人，前k个是好人，后k个是坏人。让你求一个m值，从1开始数数，只把后面的k个坏人干掉。

题目分析：

这道题因为数据量比较小。所以可以采用暴力枚举的方式求得各个k对应的m值。也可以每次算出来以后记录下来当前k对应的m值。(如果不这样做会TLE)

代码如下：

/*
 * b.cpp
 *
 *  Created on: 2015年2月2日
 *      Author: Administrator
 */

#include <iostream>
#include <cstdio>


using namespace std;
const int maxn = 15;
int ans[maxn];
int joseph[maxn];

int main(){
	int k;
	while(scanf("%d",&k)!=EOF,k){
		if(joseph[k] != 0){//如果之前已经遇到过相同的k值
			printf("%d\n",joseph[k]);//则直接输出。否则会TLE
			continue;
		}

		int m = 1;//最小间隔是1
		int n = 2*k;//计算总人数

		int i;
		for(i = 1 ; i <= k ; ++i){//判断一个m值在k轮里面是否会杀掉好人
			ans[i] = (ans[i-1]+m-1)%(n-i+1);//计算下一个要杀掉的位置.(ans[i-1]+m-1)是下一个要杀掉的索引.(n-i+1):当前剩余人数
			if(ans[i] < k){//如果杀掉了好人
				i=0;//说明当前m值不合适,从心开始试验
				m++;//去下一个m值
			}
		}
		joseph[k] = m;//额k如果已经求出m值了就记录下来。下一次如果是相同的k就不需要再次计算了

		printf("%d\n",m);
	}

	return 0;
}