用链表解决淘汰出局问题(c++)

有12个人围成一圈,标号1~12,从其中一人开始,按顺时针方向1至K报数,报到K的人被淘汰出局,再从被淘汰者后面第一人开始同样报数,报到K者同样被淘汰,这样一直报下去….输出每轮淘汰的人的序号和后剩下的人的序号数
下面是我写的程序:

#include<iostream>
using namespace std;
class node
{
public:
    int num;
    node*next;
};
void print(node*h, node*t)
{
    node*curr = h;
    for (; curr != t; curr = curr->next)
    {
        cout << curr->num << " ";
    }
    cout << t->num <<endl;
}
int main()
{
    node*head, *tail, *temp;
    head = new node;
    head->num = 1;
    tail = head;
    for (int i = 2; i <= 12; i++)//////////////////////////////创建环形,记下来好了
    {
        temp = new node;
        temp->num = i;
        temp->next = head;
        tail->next = temp;
        tail = temp;
    }
    cout << "Data";
    print(head, tail);
    int k;
    cin >> k;
    int count = 0;
    while (count < 11)
    {
        for (int i = 1; i < k;i++)//////////////////////一直数到下一个被删除的,因为是环,所以不会出错。
        {
            temp = temp->next;
        }
        cout << temp->next->num<< "=>";
        if(temp->next->num == head->num)
        {
            head = temp->next->next;
        }
        if(temp->next->num == tail->num)/////////////////如果头尾被删除,头向后移,尾向前移!!!!!!!!!!!!
        {
            tail = temp;
        }
        temp->next = temp->next->next;
        if (head ==tail)////////////////////////////////////只有一个时,break,否则一直输出最后一个数!!!!!
        {
            cout << head->num;
            break;
        }
        node*curr = head;
        for (; curr != tail;curr = curr->next)
        {
            cout << curr->num << " ";
        }
        cout << tail->num <<endl;
        count++;
    }

}

猜你喜欢

转载自blog.csdn.net/qq_38402294/article/details/71378612