使用链表实现Josephus环问题

分析：先创建一个有total个结点的链表，然后头尾相连，构成一个环形链表。从第一个结点开始数到第m个结点，从链表中删除对应结点，表示小孩出圈。然后再从被删除结点的下一个结点重新开始计数，直到链表中剩下最后一个结点。

#include  < stdlib.h >
#include  < stdio.h >
#define  LEN sizeof(struct child)
struct  child 
{
     int  num;
     struct  child  * next;
};

void  main()
{
     struct  child  * create( int  num);
     int  count( struct  child  * head, int  total, int  m);
     struct  child  * head;
     int  total,m,n;
    printf( " please input total and start: " );
     do  
    {
        scanf( " %d%d " , & total, & m);
    }  while (total < 2 || m < 2 || m > total);
    head = create(total);
    n = count(head,total,m);
    printf( " \nThe left child is %d\n " ,n);
}

struct  child  * create( int  num)
{
     struct  child  * p1, * p2, * head;
     int  i;
    p1 = ( struct  child  * )malloc(LEN);
    p1 -> num = 1 ;
    head = p1;
     for  (i = 2 ;i <= num;i ++ )
    {
        p2 = ( struct  child  * )malloc(LEN);
        p2 -> num = i;
        p1 -> next = p2;
        p1 = p2;
    }
    p1 -> next = head;                     // 头尾相连
     return  head;
}

int  count( struct  child  * head, int  total, int  m)
{
     struct  child  * p = head, * old;
     int  i,j;
     for  (i = 1 ;i < total;i ++ )             // 循环次数
    {
         for  (j = 1 ;j < m;j ++ )             // 循环个数
        {
            old = p;
            p = p -> next;
        }
        printf( " %3d " ,p -> num);
        old -> next = p -> next;
        free(p);
        p = old -> next;
    }
     return  p -> num;
}