"Brush Quiz" with an array of analog Joseph ring

Problem Description:

N Personal siege One lap, the 1th person from 1 began to count off, counted to M of the people out, and then from the next person began to count off from 1, repeat the game n-1 times, every time the elimination of 1 people, the last remaining person is the winner.

Solid method:

This is the Joseph ring problem, can be implemented with a linked list, I am here with an array implementation, the specific analysis is as follows:

Use variables: Take n=8,m=4 as an example to illustrate

R: Control cycle Times

I: Used to record the person position of each dequeue, change from 1 to 8 and then to 1

J: Used to control the count off, from 1 to 4, in to 1 ....

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Note the place:

1) Simulation of the process, starting from 1 to count, up to 4, and starting from 1, so to take the operation of 4, but the range of j%4 from 0~3, the scope of the hope is 1~4, so the correct expression of J is j=j%4+1;

2) The variable i indicates which person is reported by the report, each continued to count should be 8 after the touch plus 1, that is, i=i%8+1;

3) in the process of making a count, to skip the already out of the person, the implementation of the array value is set to its subscript, each row out of a position, the position of the element will be changed to 0, if a position corresponding to the array element value is 0, the change position is skipped. The position where the array element value is not 0 after 7 rounds is winner

#include <stdio.h> #include <stdlib.h>int main () {const int n=8;    const int m=4;    int a[9]={0,1,2,3,4,5,6,7,8};    int r=1;    int i,j;        for (i=1,j=1;r<=7;)    {while (a[i]==0) i=i%n+1;         if (j%m==0)         {printf ("a[%d]=%d\n", I,a[i]);            a[i]=0;            j=0;            r++;         }         i=i%n+1;         j=j%m+1;    }    for (i=0;i<n;i++)    {if (a[i+1]!=0) printf ("The last winner is:%d\n", a[i+1]);    } return 0;}

"Brush Quiz" with an array of analog Joseph ring