Shulou(Shulou.com)06/01 Report--

Joseph

Time Limit: 1000MS

Memory Limit: 10000KTotal Submissions: 53862

Accepted: 20551Description

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 340Sample Output 530Source Central Europe 1995 题意：同样的约瑟夫问题，只是现在有K个好人，K个坏人，确定一个步长，是的在第一个好人被清除之前所有坏人都被清除干净。 题解： 1.坏人被清除掉的先后顺序无关紧要，知道下一个除掉的是好人还是坏人就行了。2.由于好人一直都是k个，每除掉一个坏人，坏人数-1，所以队列的总数每次-1但是好人一直是前k个3.下一个被清除的人在队列中的"相对"序号为 s = (s+m-1)%(n-i); 4.只要s>

=k, then the next one to be eliminated is the bad guy 5. Next, let's talk about the value range of m: Let's examine the situation when there are only k+1 people left, that is, there is still one bad guy who has not been executed, so in this round, the end position must be at the last bad guy, so where is the starting position? This requires finding the end position of K+2 people, however, the end position of K+2 people must be K+2 people or K+1 people, so there are two order cases: GGGG... GGGXB or GGGG... GGGBX (X represents the person who exits the round with K+2 people) So there are two possibilities for the starting position of the round with K+1 people, namely, the first position or the position of K+1. There are two possibilities for limiting m: t(k+1) or t(k+1)+1; t>=1; if traversing every m must timeout, it is necessary to make a table and limit the range of m to avoid timeout.

AC codes are given below:

1 #include 2 //#include 3 using namespace std; 4 int a[14]; 5 int f(int k,int m) 6 { 7 int n,i,s; 8 n=2*k; 9 s=0;10 for(i=0;i

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