pku1012.txt

北京大学 在线acm在线判题系统 一些题目解答

Joseph 
Time Limit:1000MS  Memory Limit:10000K
Total Submit:10607 Accepted:3964 

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



Source
Central Europe 1995

Source

/*
   农夫三拳@seu
   drizzlecrj.gmail.com
   2007-03-02
*/
Problem Id:1012  User Id:Drizzle 
Memory:36K  Time:421MS
Language:C++  Result:Accepted

Source 

#include <stdio.h>

int MinTime(int k)
{
	int i, j, m, m2, n, r;
	n = k + k;
	m = k ;
	while(true)
	{
		++m;
		j = 0;
		r = n;
		while(r > k)
		{
			m2 = (m - 1) % r + 1;
			j = (j + m2 - 1) % r;
			if(j < k)
				break;
			r--;
		}
		if(r == k)
			return m;
	}
}

int main()
{
	int res[14];
	int k;
	for(int i = 1; i < 14; i++)
		res[i] = MinTime(i);
	while(scanf("%d", &k) && k)
	{
		printf("%d\n", res[k]);
	}

	return 0;
}