数据结构1 (链表) 约瑟夫

最新推荐文章于 2022-04-06 21:24:55 发布

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最新推荐文章于 2022-04-06 21:24:55 发布

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分类专栏：学习

本文链接：https://blog.youkuaiyun.com/SakuraZhangrx/article/details/111145766

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本文介绍了一个基于约瑟夫环问题的实现方案，采用链表和数组两种方式来模拟这一经典问题。通过输入每个人的编号和初始报数大小，程序能够计算并输出最后剩余的人的编号。

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#include <iostream>
#include<algorithm>
using namespace std;

const int N = 10010;
int a[N];
int n, firstm;
struct node
{
	int num;
	int password;
	struct node *next;

};
 
class joseph
{
public:
	node* Creatnode(int i, int password)
	{
		node* p;
		p = (node*)malloc(sizeof(node));
		p->num = i;
		p->password = password;
		p->next = NULL;
		return p;
	}
	node* Creatjoseph(int n)
	{
		node* head=NULL, * p=NULL, *q=NULL;
		for (int i = 1; i <= n; i++)
		{
			p = Creatnode(i, a[i]);
			if (i == 1)
				head = p;
			else
				q->next = p;
			q=p;
		}
		q->next = head;
		node* it = head;
		for (int i = 1; i <= n; i++)
		{
			//cout << it->num << " " << it->password << endl;
			it = it->next;
		}
		//cout << it->num << endl;
		return head;
	}
	void Runjoseph(int n, int firstm)
	{
		node *p, *q=NULL;
		p = Creatjoseph(n);
		for (int i = 1; i < firstm-1; i++)
		{
			p = p->next;
		}
		q = p->next;
		p->next = q->next;
		p = p->next;
		cout << q->num << " ";
		int temp = q->password;
		//cout << temp << " ";
		free(q);
		while (p->next!=p)
		{
			if (temp > 1)
			{
				for (int i = 1; i < temp - 1; i++)
				{
					p = p->next;
				}
			}
			else
			{
				p = p->next;
			}
			q = p->next;
			p->next = q->next;
			p = p->next;
			cout << q->num << " ";
			temp = q->password;
			//cout << temp <<" ";
			free(q);
		}
		cout << p->num << endl;
	}
};
int x[N];
int b[N];
void runjosephb(int x[], int m)
{
	int k = 1, i = 0, cnt = a[0], count = 1 , l = n;
	for (;l >= 1;)
	{
		if (count==cnt)
		{
			k = i % n;
			x[k] = 1;
			printf("%d ",k+1);
			l--;
			count = 0;
			cnt = a[i%n+1];
		}
		i++;
		if (x[i % n] != 1)
			count++;
	}
}
int main()
{
	joseph zrx;
	cout << "请输入长度和第一个人报数出圈大小" << endl;
	cin >> n >> firstm;
	cout << "请输入每个人所带的数大小" << endl;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	for (int i = 1; i <= n; i++)
	{
		b[i] = a[i];
	}
	a[0] = firstm;
	int c;
	cout << "请输入构造方式？ 1:数组,2:链表";
	cin >> c;
	if (c == 2)
		zrx.Runjoseph(n, firstm);
	else
	{
		/*for (int i = 1; i <= n; i++)
		{
			a[i - 1] = b[i];
		}*/
		for (int i = 0; i < n; i++)
			x[i] = 0;
		runjosephb(x,firstm);
	}
	//zrx.Creatjoseph(n);
	return 0;
}