约瑟夫生死问题

约瑟夫环生死问题,利用循环链表作为存储结构,使用简单的c++语言知识便可实现。

由于算法并不复杂,故并未使用多个函数,在一个主函数中实现

#include<iostream>
#include<cstdlib>
using namespace std;
typedef int ElemType;
typedef struct lnode
{
	ElemType data;
	struct lnode *next;
}node, *linklist;
int main()
{
	int n, i = 1, k;
	node *p, *q;
	linklist head;
	cin >> n >> k;
	if (n % 2 != 0)
		cout << "请输入偶数,否则无法实现。" << endl;
	head = (node*)malloc(sizeof(node));
	head->data = 1;
	q = head;
	p = head;
	for (int l = 2; l <= n; l++)
	{
		p = (linklist)malloc(sizeof(node));
		q->next = p;
		q = p;
		p->data = l;
	}
	q->next = head;//循环单链表创建完成
	p = head;
	cout << "死亡人员:";
	while (i <= n / 2)
	{
		for (int m = k; m >2; m--)
			p = p->next;
		q = p->next;
		p->next = q->next;
		cout << q->data << " ";
		free(q);
		p = p->next;
		i++;
	}//死亡人员确认
	cout << endl << "幸存人员:";
	i = 1;
	while (i <= n / 2)
	{
		p = p->next;
		cout << p->data << " ";
		i++;
	}
	cout << endl;
	return 0;
}
    原文作者:约瑟夫环问题
    原文地址: https://blog.csdn.net/zarkpx/article/details/53573738
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
点赞