双向约瑟夫--双向循环链表

WA了无数次..最终搞懂题意..然后修改代码= =

思路很简单的...自己基础薄0 0

丑陋代码贴上= =

#include<stdio.h>
#include<stdlib.h>
#define MALLOC (STU*)malloc(sizeof(STU))
int m1, m2, n;
typedef struct person{
    struct person *pre, *next;
    int lis;
}STU;

void _delete(struct person *node)
{
    node->pre->next = node->next;
    node->next->pre = node->pre;
    free(node);
}
int main()
{
    int i;
    STU *head = MALLOC, *temp, *_pre;
    scanf("%d%d%d", &n, &m1, &m2);
    head->lis = 1;  _pre = head;
    for(i = 2; i <= n; i++)//temp为当前申请结点,_pre为前一个结点,结束之后_pre当尾结点用.
    {
        temp = MALLOC;
        temp->pre = _pre;
        temp->lis = i;
        _pre->next = temp;
        _pre = temp;
    }
    //链接首尾
    temp = head->pre = _pre;
    _pre->next = head;
    while(n > 2)//只剩下一个结点结束
    {
        for(i = 0; i < m1; i++) temp = temp->next;
        _pre = temp->next; //存放删除结点
        _delete(temp);
        temp = _pre;
        if(temp == temp->next) break;
        for(i = 0; i < m2; i++) temp = temp->pre;
        if(_pre == temp) { _pre = temp->pre; _delete(temp); temp = _pre; n -= 1; continue; }
        _pre = temp->pre;
        _delete(temp);
        temp = _pre;
        n -= 2;
    }
    printf("%d\n", temp->lis);
    return 0;
}