约瑟夫经典问题扩展成双向约瑟夫问题

#include<stdio.h> 
#include<malloc.h> 
#include<stdlib.h> 
typedef int ElemType; 
int n,m,k; //定义一个全局变量  
typedef struct DuLNode    //双向循环链表结构 
{ 
  ElemType data; 
  struct DuLNode *prior; 
  struct DuLNode *next; 
}DuLNode,*DuLinkList;   
void Create(DuLinkList &H)  //创建带头结点的双向循环链表  
{ 
  DuLinkList p,q; 
  int i; 
  H=(DuLinkList)malloc(sizeof(DuLNode));  
  p=H; 
  q=H; 
  for(i=1;i<=n;i++) 
  { 
    p=(DuLinkList)malloc(sizeof(DuLNode)); 
    p->data=i; 
    p->prior=q; 
    q->next=p; 
    q=p; 
  } 
  p->next=H; 
  H->prior=p; 
} 
void Delete(DuLinkList &P) //删除结点  
{ 
  P->prior->next=P->next; 
  P->next->prior=P->prior; 
} 
int main() 
{ 
  int i; 
  DuLinkList H,l,R,right,left; //分别用以表示头结点,l和R都用于找k的值，向右（顺时针），向左（逆时针）  
  scanf("%d,%d,%d",&n,&k,&m);  
    if(!n||!k||!m) 
    { 
      printf("n,m,k must bigger than 0.\n"); 
      return 0;   
    } 
    if(k>n) 
    { 
      printf("k should not bigger than n.\n"); 
      return 0; 
    }     
    Create(H); 
    R=H->next ; 
    while(R->data!=k) 
    { 
      R=R->next; 
    } 
    l=R; 
    while(n) 
    { 
      right=R; 
      left=l; 
      for(i=1;i<m;i++) 
      { 
        right=right->next; 
        left=left->prior; 
        //遇见头结点需要特殊处理 
        if(right==H) 
          right=right->next; 
        if(left==H) 
          left=left->prior; 
      } 
      R=right->next; 
      l=left->prior; 
      if(R==H) 
        R=R->next; 
      if(l==H) 
        l=l->prior; 
      if(right!=left) 
      { 
        n=n-2; 
        printf("%d-%d,",right->data,left->data); 
        Delete(right); 
        Delete(left); 
      } 
      else 
      { 
        n=n-1; 
        printf("%d,",right->data); 
        Delete(right);  
      } 
    } 
    printf("\n");  
}