【数据结构作业—01】用单循环链表解决约瑟夫问题

实验作业一：线性表（链表）

1. 用单循环链表解决约瑟夫问题。

问题描述：

一个旅行社要从n个旅客中选出一名旅客，为他提供免费的环球旅行服务。旅行社安排这些旅客围成一个圆圈，从帽子中取出一张纸条，用上面写的正整数m(<n)作为报数值。游戏进行时，从第s个人开始按顺时针方向自1开始顺序报数，报到m时停止报数，报m的人被淘汰出列，然后从他顺时针方向上的下一个人开始重新报数，如此下去，直到圆圈中只剩下一个人，这个最后的幸存者就是游戏的胜利者，将得到免费旅行的奖励。其中数据结构采用单循环链表。

解决方案要求：

输入参数：n、m、s

输出参数：n个人的淘汰序列

参考样例：

 

代码：

#include "stdlib.h"
#include "stdio.h"
#include <iostream>
using namespace std;

typedef int datatype;

typedef struct node {
    datatype data;
    struct node *next;
}node, *LinkList;

void Inite(LinkList &first, int n)    {
    first = (node*)malloc(sizeof(node)); //注意此处，不可以直接分配长度为n*sizeof(node)的空间，因为这里只可以给头节点分配，如果分配多了也没用
    node *p = first, *q;
    //first->data = 1;
    //cout << first->data;
    for(int i = 0; i < n - 1; i++)    {
        p->data = i + 1;
        q = (node*)malloc(sizeof(node));
        //cout << "Number " << i+1 << " p->data  " << p->data << endl;
        p -> next = q;
        p = q;
    }
    p -> data = n;
    p -> next = first;
    //cout << "Number " << n << " p->data  " << p->data << endl;
}

void Josephus(LinkList &first, int m, int s)    {
    cout << "******** Solve Josephus Problem ********" << endl;
    
    node *nowPoint = first, *prePoint = first;
    if (s > 1)    {
        for (int i = 0; i < s - 1; i++)    {
            prePoint = nowPoint;
            nowPoint = nowPoint -> next;
            //cout << "NUMBER " << i + 1 << " prePoint " << prePoint->data << endl;
            //cout << "NUMBER " << i + 1 << " nowPoint " << nowPoint->data << endl;
        }
    }
    else if(s == 1)    {
        while(prePoint -> next != nowPoint)
            prePoint = prePoint -> next;
    }
    else    {
        printf("PLEASE ENTER AN S WHICH BIGGER THAN 4 !");
    }
    while (nowPoint -> next != nowPoint)    {
        for (int i = 0; i < m - 1; i++)    {
            prePoint = nowPoint;
            nowPoint = nowPoint -> next;
            //cout << "NUMBER " << i + 1 << " prePoint " << prePoint->data << endl;
            //cout << "NUMBER " << i + 1 << " nowPoint " << nowPoint->data << endl;
        }
            prePoint -> next = nowPoint -> next;
            cout << "Number      " << nowPoint -> data << " is out" << endl;
            free(nowPoint);
            nowPoint = prePoint -> next;
    }
    cout << "Number      " << nowPoint -> data << " is out" << endl;
    cout << "****************** END *****************" << endl;
}

int main()    {
    int n, m, s;
    LinkList first;
    
    cout << "Enter n:" << endl;
    cin >> n;
    cout << "Enter m:" << endl;
    cin >> m;
    cout << "Enter s:" << endl;
    cin >> s;
    
    Inite(first, n);
    Josephus(first, m, s);
    
    return 0;
}