HDU Joseph【数学&&约瑟夫环】

Joseph

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2118 Accepted Submission(s): 1285

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

以前做过一次约瑟夫环问题，不过没太明白当时大神的最优代码，这一次又见到了类似的题，只能无脑模拟了，不过暴力模拟，很果断的TLE了.......

找大神们的解析，加深了自己的理解，不过自己还是不太明白，貌似是，利用数学技巧，动态处理了，简化了很多东西...（自己说不明白了）

然后就是打表处理了.....

#include<stdio.h> #include<string.h>bool joseph(int k,int m){int bg=0,ed=k-1,tp;for(int n=2*k;n>k;--n){tp=(m-1)%n;if(tp>=bg&&tp<=ed){return 0;}bg=(bg-m%n+n)%n;ed=(ed-m%n+n)%n;//bg=((bg-m)%n+n)%n;ed=((ed-m)%n+n)%n;//同余还是理解的不够}return 1;}int main(){int n,y[20]={0};for(int i=1;i<14;++i){int j=i+1;while(1){if(joseph(i,j)){y[i]=j;break; }++j;} }//freopen("shuju.txt","r",stdin);while(scanf("%d",&n),n){printf("%d\n",y[n]);}return 0;}