约瑟夫问题

/* * 说有15个教徒和15个非教徒在海上遇难，必须将一半的人投入海中，其余的人才能幸免于难。于是他们想了一个办法：30个人围成一圈，从第一个人开始报数，每数到第九个人的时候将之投入大海，如此循环直到剩下15个人为止。问：怎样的排法可以使被投入大海的都是非教徒？ * * StoryMonster    2016/7/14 */#include <iostream>using namespace std;enum {BELIEVER=0,NONBELIEVER};int main(){    unsigned short people[30] = {BELIEVER};    unsigned short surviver = 30;    unsigned short index = 0;    while(surviver > 15)    {        unsigned short count = 0;        while(1)        {            if(people[index] == BELIEVER)            {                count++;                if(count == 9) break;            }            index++;            if(index == 30) index = 0;        }        people[index] = NONBELIEVER;        surviver--;    }    for(index=0;index<30;index++)    {        if(people[index] == BELIEVER)            cout<<"BELIEVER  ";        else cout<<"NONBELIEVER  ";    }    cout<<endl;    return 0;}