POJ 1012

参考自：https://www.cnblogs.com/ECJTUACM-873284962/p/6480880.html

Joseph

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 53862		Accepted: 20551

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

分析：

　　坏人被清除掉的先后顺序无关紧要，知道下一个除掉的是好人还是坏人就行了。由于好人一直都是k个，每除掉一个坏人，坏人数就-1，队列的总数就每次-1，而下一个被清除的人在队列中的“相对”序号为 s = (s+m-1)%(n-i)。只要s>=k，那么下一个被除掉的就是坏人。

　　接下来说说m的取值范围：我们考察一下只剩下k+1个人时候情况，即坏人还有一个未被处决，那么在这一轮中结束位置必定在最后一个坏人，那么开始位置在哪呢？

　　这就需要找K+2个人的结束位置，然而K+2个人的结束位置必定是第K+2个人或者第K+1个人，这样就出现两种顺序情况：GGGG.....GGGXB 或  GGGG......GGGBX (X表示有K+2个人的那一轮退出的人）所以有K+1个人的那一轮的开始位置有两种可能即第一个位置或K+1的那个位置。

　　限定m有两种可能： t(k+1) 或 t(k+1)+1; t>=1; 若遍历每一个m必定超时，避免超时则需要打表和限制m的范围。

代码：

#include <bits/stdc++.h>
using namespace std;
int f(int k,int m){//死亡判断 8         s=(s+m-1)%(n-i);
        if(s<k)return 0;
    }
    return 1;
}
int main(){
    int i,k,n,a[14];
    for(k=1;k<=14;k++){
        i=k+1;
        while(1){
            if(f(k,i)){
                a[k]=i;
                break;
            }else if(f(k,i+1)){
                a[k]=i+1;
                break;
            }
            i+=k+1;//再加一倍的k+1
        }
    }
    while(scanf("%d",&n)&&n)cout<<a[n]<<endl;
    return 0;
}