【數據結構作業—01】用單循環鏈表解決約瑟夫問題

實驗作業一：線性表（鏈表）

1. 用單循環鏈表解決約瑟夫問題。

問題描述：

一個旅行社要從n個旅客中選出一名旅客，為他提供免費的環球旅行服務。旅行社安排這些旅客圍成一個圓圈，從帽子中取出一張紙條，用上面寫的正整數m(<n)作為報數值。游戲進行時，從第s個人開始按順時針方向自1開始順序報數，報到m時停止報數，報m的人被淘汰出列，然后從他順時針方向上的下一個人開始重新報數，如此下去，直到圓圈中只剩下一個人，這個最后的幸存者就是游戲的勝利者，將得到免費旅行的獎勵。其中數據結構采用單循環鏈表。

解決方案要求：

輸入參數：n、m、s

輸出參數：n個人的淘汰序列

參考樣例：

 

 

代碼：

#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)); //注意此處，不可以直接分配長度為n*sizeof(node)的空間，因為這里只可以給頭節點分配，如果分配多了也沒用
    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;
}