POJ 1012 Joseph

Joseph

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 53862		Accepted: 20551

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

題意：一樣的約瑟夫問題，只是如今有K個好人，K個壞人，肯定一個步長，是的在第一個好人被清除以前全部壞人都被清除乾淨。

題解：

 

1.壞人被清除掉的前後順序可有可無，知道下一個除掉的是好人仍是壞人就好了。
2.因爲好人一直都是k個，每除掉一個壞人，壞人數-1，因此隊列的總數每次-1可是好人一直是前k個
3.下一個被清除的人在隊列中的「相對」序號爲 s = (s+m-1)%(n-i);
4.只要s>=k，那麼下一個被除掉的就是壞人

5.接下來講說m的取值範圍：咱們考察一下只剩下k+1我的時候狀況，即壞人還有一個未被處決，那麼在這一輪中結束位置一定在最後一個壞人，那麼開始位置在哪呢？這就須要找K+2我的的結束位置，然而K+2我的的結束位置一定是第K+2我的或者第K+1我的，這樣就出現兩種順序狀況：GGGG.....GGGXB 或  GGGG......GGGBX (X表示有K+2我的的那一輪退出的人）因此有K+1我的的那一輪的開始位置有兩種可能即第一個位置或K+1的那個位置，限定m有兩種可能：t(k+1) 或 t(k+1)+1; t>=1; 若遍歷每個m一定超時，避免超時則須要打表和限制m的範圍。

下面給出AC代碼：

 

 1 #include <cstdio> 2 //#include <bits/stdc++.h> 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<k;i++)11     {12         s=(s+m-1)%(n-i);13         if(s<k) return 0;14     }15     return 1;16 }17 int main()18 {19     int i,k,n;20     for(k=1;k<=14;k++)21     {22         i=k+1;23         while(1)24         {25             if(f(k,i))26             {27                 a[k]=i;28                 break;29             }30             else if(f(k,i+1))31             {32                 a[k]=i+1;33                 break;34             }35             i+=k+1;36         }37     }38     while(scanf("%d",&n)&&n)39     printf("%d\n",a[n]);40     return 0;41 }