POJ 2244 Eeny Meeny Moo (约瑟夫环问题)

Eeny Meeny Moo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2929		Accepted: 2015

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

3

4

5

6

7

8

9

10

11

12

0

Sample Output

2

5

2

4

3

11

2

3

8

16

Source

Ulm Local 1996

 

题意 ：N个城市轮流断电，先把城市1断掉，之后便剩下n-1个城市，在这种情况下，求最小的M使胜者为最初的城市2 (换句话说就是现在的城市1)。现在将问题转化一下：在 1...n-1 个数中，求最小的数m，使得胜者为1。
算法 ：数学递推
参考 ： http://www.cppblog.com/AClayton/archive/2007/11/06/35964.html

 

#include<stdio.h>



int main(){



    //freopen("input.txt","r",stdin);



    int n;

    while(~scanf("%d",&n) && n){

        int m,s;

        for(m=1;;m++){

            s=0;

            for(int i=2;i<n;i++)

                s=(s+m)%i;

            if(s==0){

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

                break;

            }

        }

    }

    return 0;

}