Joseph
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3428 Accepted Submission(s): 1989
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
瞎胡翻译：有一圈人，前半圈k个好人，后半圈k个坏人，若总人数n=6，m=5，执行的顺序是5, 4, 6, 2, 3，剩下的一个会存活。要求先处决完坏人，再处决好人，求m。
蜜汁思路：刚开始一直不理解题目的要求，从1开始把第m个人处决，然后从m+1的位置开始再查m个人，把第m个处决，重复这个过程，直到只剩下一个人。没有一个好的思路，只能先暴力，虽然k不大，但是暴力也会超时。最后，把14个结果手动打表。
#include<stdio.h>
int main()
{
int k;
int ans[14]={0, 2,7,5,30,169,441,1872,7632,1740,93313,459901,1358657,2504881};
while(scanf("%d",&k)!=EOF&&k>0)
{
printf("%d\n",ans[k]);
}
}