acm hnu 134 Joseph's Puzzle

最新推荐文章于 2022-07-30 16:20:03 发布

原创最新推荐文章于 2022-07-30 16:20:03 发布 · 1.8k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#integer #each #input #express #output #numbers

Algorithm 专栏收录该内容

33 篇文章

订阅专栏

本文介绍了一个经典的约瑟夫环问题，通过设定初始人数及计数规则，模拟了淘汰过程，并提供了具体的输入输出样例与实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Joseph's Puzzle

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB

Total submit users: 134, Accepted users: 125

Problem 10045 : No special judgement

Problem description

Persons, whose number order is from 1 to N, hold a password (the straight integer less than 10000) each person and sit around clockwise.
At first we choose a positive integer factor as the count off number M and start to number clockwise from number one .As we number to M the person who numbers M leaves the rank and whose password will be the new number M. And then start to number again from the person who is sit clockwise next to the person who has been out off the rank just before then. Continue like this, until all people are out off the rank completely. Please design a program of leaving order.

Input

There have some groups data. First line of each group data is two integers N, M (0<N, M<100),the second line is N positive integer, separately express every person’s password.The last line contins 0,0 ,which means the end of input data.

Output

To each group of data, according to the order which leaves outputs their serial number. Between the numeral separates with a blank space. Each group of data monopolize line of outputs.

Sample Input

7 20
3 1 7 2 4 8 4
4 3
1 2 3 4
0 0

Sample Output

6 1 4 7 2 3 5
3 2 1 4

Problem Source

HNU 1'st Contest

/*18:10~  */

#include <stdio.h>

#define MAX 50000

int move( int num[],int Mr );

main()

{

      int num[MAX],n,Mright=0,p=0,total=0,M=0,Mr=0;

      int i,j;

      i = 0;

      for (i = 0 ; i < MAX ; i++)num[i]=-2;


      scanf("%d%d",&n,&M);

      for (i = 0 ; i < n ; i++)scanf("%d",&num[i]);

      while( n!=0 && M!=0 ){

             total = n;



             p = 0;

             Mr = M%n;

             while( total > 0 ){

                    p=move(num,Mr);

            if(total==1){printf("%d ",p+1);break;}

            else  printf("%d ",p+1);

            M = num[p];//record M

            num[p]=-1;

            total--;

            p++;    //delete it

                    M=M%total;

            //caculate the amount after this

            i=p;

            Mright=0;

            while(num[i]!=-2){

              if(num[i]!=-1)Mright++;

              i++;

            }

            if(M==0&&total==1){Mr=1;continue;}

            if(M==0)Mr=total-Mright;

            else if(M>Mright)Mr=M-Mright;

            else Mr=total+M-Mright;

          }//while total

              for (i = 0 ; i < MAX ; i++)num[i]=-2;

              scanf("%d%d",&n,&M);
              for (i = 0 ; i < n ; i++)scanf("%d",&num[i]);

      }

return 0;
}

//return current p,M < total

int move( int num[],int Mr )

{

    int c=0,p=0;

    while(1){

             if( num[p] != -1 ){

                 c++;

                if(c == Mr)return p;

                 }

                 p++;

             }

}