Eeny Meeny Moo（经典Josephus变体（三大模板））

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本文探讨了在特定条件下的城市网络接入问题，通过引入随机数m和城市总数n，提出了一个算法来确保德国某城市始终能保持网络连接。算法通过解决Josephus问题的变体，确保了该城市作为最后一个被排除在外的节点。文章详细阐述了解题思路，并提供了多个版本的代码实现，以提高效率。

Link：http://poj.org/problem?id=2244

Eeny Meeny Moo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3203		Accepted: 2219

Description

Surely you have made the experience that when too many people use the Internet simultaneously, the net becomes very, very slow.
To put an end to this problem, the University of Ulm has developed a contingency scheme for times of peak load to cut off net access for some cities of the country in a systematic, totally fair manner. Germany's cities were enumerated randomly from 1 to n. Freiburg was number 1, Ulm was number 2, Karlsruhe was number 3, and so on in a purely random order.
Then a number m would be picked at random, and Internet access would first be cut off in city 1 (clearly the fairest starting point) and then in every mth city after that, wrapping around to 1 after n, and ignoring cities already cut off. For example, if n=17 and m=5, net access would be cut off to the cities in the order [1,6,11,16,5,12,2,9,17,10,4,15,14,3,8,13,7]. The problem is that it is clearly fairest to cut off Ulm last (after all, this is where the best programmers come from), so for a given n, the random number m needs to be carefully chosen so that city 2 is the last city selected.

Your job is to write a program that will read in a number of cities n and then determine the smallest integer m that will ensure that Ulm can surf the net while the rest of the country is cut off.

Input

The input will contain one or more lines, each line containing one integer n with 3 <= n < 150, representing the number of cities in the country.
Input is terminated by a value of zero (0) for n.

Output

For each line of the input, print one line containing the integer m fulfilling the requirement specified above.

Sample Input

Sample Output

Source

Ulm Local 1996

解题思想：有n个城市，第一个总是最先退出，相当于从第二个开始按1,2，……m，1,2……m这样报数，问题实际上可抽象成n-1个城市的Josephus问题，注意josephus思想是按0,1，……n-1编号的，又加上是从第二个城市开始报数，求得的最后一个城市的实际编号=Josephus方法求得的编号 +1+1。

效率较低版本：

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
	int i,j,m,f,a[200];
	for(i=3;i<=150;i++)
	{
		f=0;
		m=1;
		while(1)
		{
			for(j=2;j<=i-1;j++)
			{
				f=(f+m)%j;
			}
			if(f+1+1==2)
			{
				a[i]=m;
				break;
			}
			else
			{
				m++;
				f=0;
    			}
		}
	}


	int n;
	while(scanf("%d",&n)==1&&n)
	{
		printf("%d\n",a[n]);
	}
	return 0;
}

效率较高版本：

#include<iostream>
#include<algorithm>
using namespace std;
int josephus(int n,int m,int s)
{
	if(m==1)
		return (s+n-1)%n;
	for(int i=2;i<=n;i++)
	{
		s=(s+m)%i;
		if(i==n)
		break;
		if(s+m<i)
		{
			int x=(i-s)/(m-1);
			if(i+x<n)
			{
				i=i+x;
				s=(s+m*x)%i;
			}
			else
			{
				s=(s+m*(n-i))%n;
				break;
			}
		}
	}
	return s;
}
int main()
{
	int a[155],i,m,ret,n;
	for(i=2;i<=150;i++)
	{
		for(m=2;;m++)
		{
			ret=josephus(i,m,0);
			if(ret+1+1==2)
			{
				a[i]=m;
				break;
			}
		}
	}
	while(scanf("%d",&n)==1&&n)
	{
		printf("%d\n",a[n-1]);
	}
	return 0;
}

版本3：

解析：

递推公式为：

编号从0,1，……n-1，报数从1，2，……m的条件下：

f[i]; //第i轮杀掉对应当前轮的编号为f[i]的人

f[0]=0;

f[i]=(f[i-1]+m-1)%(n-i+1); (i>1 , 总人数n，则n-i+1为第i轮的人数)

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
	int n,m,i,ans[200],f;
	for(n=2;n<=150;n++)
	{
		f=0;
		m=1;
       		for(i=1;i<=n-1;i++)//经过n-1轮后剩下最后一个
       		{
    			f=(f+m-1)%(n-i+1);

    			if(f+1+1==2)
    			{
  			  	i=0;
    				m++;
			}
    		}
    		ans[n]=m;//剩下的最后一个城市为原来的第二个
	}
	while(scanf("%d",&n)==1&&n)
	{
		printf("%d\n",ans[n-1]);
	}
	return 0;
}