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
340
Sample Output
530 题目大意： 有k个好人和k个坏人，从第一个人开始循环报数（即报到m后从1开始报），报到m的人被处死，要求全部的坏人都处死，好人留下，求出m。
这个代码就是有点慢，所以我打了个表 #include using namespace std;int main(){ int k; while(cin>>k) { if(!k) break; int m=1; int ans[30]={0}; for(int i=1;i<=k;i++) { ans[i]=(ans[i-1]+m-1)%(2*k-i+1); if(ans[i]
#include int main(){ int k; int Joseph[14]={0,2,7,5,30,169,441,1872,7632,1740,93313,459901,1358657,2504881}; while(scanf("%d",&k),k) printf("%d\n",Joseph[k]); return 0;}