HDOJ 1443 joseph （找规律）

Joseph

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3036 Accepted Submission(s): 1710 

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

340

 

Sample Output

530

题意：从2k个人里面，把后面k个坏人处决掉。

解题思路：我是暴力枚举，从k+1开始，这样保证是最小的。

难点就是这个模拟过程和超时。

举例：1 2 3 4 5 6  一开始m是4，第一次执行完结果是 1 2 3 5 6 然后从 3 开始往后数m次,结果选中的数是2，2是好人，退出循环。然后m是5，第一次执行完结果是1 2 3 4 6然后从4开始数m次，结果是4，继续执行，结果是6。剩余3个结束循环。

#include  #include  #include  #includeusing namespace std;int a[20] = { 0 };int solve(int k){if (a[k])return a[k];//不这么写就超时了  因为测试数据比较多放在数组里就不会超时int c = 2 * k;//c 表示总的个数int i;while (c != k){for (i = k + 1; c != k; i++){c = 2 * k;int b = 0;//余数for (; c != k;){b = (i + b - 1) % c;//这里减一是因为去掉一个数之后  要从前一个数开始数m  if (b >= k) { c--; }//个数减一else break;//因为要从k+1个里处决人 不满足就退出去}}}return a[k] = i - 1;//减一是因为找到之后i又加了1}int main(){int b;while (cin >> b){if (b == 0)break;cout << solve(b) << endl;}return 0;}