HDU1443-Joseph

Joseph

                                                                          Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                                     Total Submission(s): 2571    Accepted Submission(s): 1529

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

 

Source

ACM暑期集训队练习赛（5）

 

题意： 有k个好人跟k个坏人按顺序坐着，然后按第m个杀人，求出把坏人全部先杀光的m的最小值。

解题思路：在这么多个人中，始终用start跟end来确定好人那个序列的位置。 比如一开始是1 2 3 4 5 6，那么start = 0,end = 3(从0开始计数)当m等于5的时候，kill后就剩下1 2 3 4 6 ，重新拍下序列变成 6 1 2 3 4 ，这时候 start = ((start-m)%n+n)%n; end =  ((end-m)%n+n)%n; n是当前剩下的人数。定完位置之后，kill的位置为kill = (m-1)%n。

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <set>#include <map>using namespace std;#define LL long longbool Find(int k,int m){    int s=0,e=k-1;    int p;    for(int i=2*k;i>k;i--)    {        p=(m-1)%i;        if(p>=s&&p<=e) return false;        s=((s-m)%i+i)%i;        e=((e-m)%i+i)%i;    }    return true;}int main(){    int x[14];    for(int i=1;i<14;i++)    {        for(int j=1;;j++)        {            if(Find(i,j))            {                x[i]=j;                break;            }        }    }    int k;    while(scanf("%d",&k)&&k)    {        printf("%d\n",x[k]);    }    return 0;}