用单向循环链表实现约瑟夫环问题

约瑟夫环的问题和种类有很多，这是一道基本的约瑟夫环问题，通过单向循环链表实现模拟此过程。

代码如下：

#include<stdio.h>
#include<stdlib.h>
typedef struct Node
{
    int num;
    int password;
    struct Node *next;
}CircleNode,*CircleList;

CircleList Create(int n);
void Knockout(CircleList L,int n,int m);

int main(void)
{
    printf("Enter the upper limit m: ");
    int m;
    scanf("%d",&m);
    printf("Enter the number of people: ");
    int n;
    scanf("%d",&n);
    CircleList L=Create(n);
    printf("The order of knockout: ");
    Knockout(L,n,m);
    return 0;
}

CircleList Create(int n)
{
    CircleList L=(CircleList)malloc(sizeof(CircleNode));
    L->next=L;
    CircleNode *rear=L;
    printf("Enter the password for person 1-%d: ",n);
    for(int i=1;i<=n;i++)
    {
        CircleNode *p=(CircleNode*)malloc(sizeof(CircleNode));
        scanf("%d",&p->password);
        p->num=i;
        rear->next=p;
        rear=p;
    }
    rear->next=L;
    return L;
}

void Knockout(CircleList L,int n,int m)
{
    CircleNode *cur=L;
    CircleNode *prev=L;
     while(n--)
    {
        for(int i=1;i<=m;)
        {
            if(cur->next!=L)
            {
                prev=cur;
                cur=cur->next;
            }
            else
            {
                prev=L;
                cur=L->next;
            }
            ++i;
        }
        printf("%d ",cur->num);
        m=cur->password;
        prev->next=cur->next;
        free(cur);
    }
}

/*
测试集
输入： m=20 n=7  密码：3 1 7 2 4 8 4
输出： 6 1 4 7 2 3 5
*/

/*
void Print(CircleList L)
{
    CircleNode *p=L->next;
    while(p!=L)
    {
        printf("%d ",p->num);
        p=p->next;
    }
}
*/

原文地址：https://www.cnblogs.com/CuteyThyme/p/10679352.html