1012-Joseph 约瑟夫问题

问题描述：

Joseph

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 52628		Accepted: 20053

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
中文意思：

题意：给出一个k，代表有K个好人与K个坏人，其中前K个是好人，后K个是坏人，根据约瑟夫环游戏的原理，在坏人都出局而好人没有一个出局的情况下，m最小是多大

思路：由于k的数值小，直接暴力枚举打表

c++代码：

#include<iostream>
#include<stdio.h>
using namespace std;
int main(){
int min, countx = 0, a[14], *T = new int[100], P[14] = { 0 };//T表示输入的各个数字
memset(a,0,sizeof(int)*14);//memset函数用于集体赋值，a表示数组的起始地址，0表示赋的值，sizeof（int）表示int的尺寸*数组的大小
for (int i = 1; i < 14; i++){
memset(a, 0, sizeof(int)* 14);
min = 1;
for (int j = 1; j <= i; j++){
a[j] = (a[j - 1] + min - 1) % (2 * i - j + 1);//a[j-1]+min-1表示上一局过后的起始位置开始数min个数到现在的位置，
//2*i-j+1表示到现在这一轮总共的人数，因为有可能超出所以要取余数
if (a[j] < i){
j = 0;
min++;
}
}
P[i] = min;
}
while (cin >> T[countx] && T[countx] != 0)countx++;
countx = 0;
while (T[countx++] != 0)
cout <<P[T[countx - 1]] << endl;
delete []T;
return 0;
}