POJ 3750 小孩报数问题

基本的双向链表即可实现。

最后一个element的判定很重要。

 

#include <iostream>
#include <string>

using namespace std;

typedef struct Child
{
    string name;
    Child* pre;
    Child* next;
} Child;


int main()
{
    // Get input N
    int n;
    cin>>n;

    Child* head = NULL;
    Child* tail = NULL;
    Child* cur = NULL;

    // Create dlink list
    for(int i=0; i<n; i++)
    {
        cur = new Child();
        cin>>cur->name;
        if(head == NULL)
        {
            head = cur;
            tail = cur;
            cur->next = cur;
            cur->pre = cur;
        }
        else
        {
            head->pre = cur;
            tail->next = cur;
            cur->pre = tail;
            cur->next = head;
            tail = cur;
        }
    }

   cur = head;
   int w,s;

   cin>>w;
   cin.get(); // ignore comma
   cin>>s;

   // get the W-th element
   if(cur != NULL)
   {
       while(w-1 != 0)
       {
           cur = cur->next;
           w--;
       }
   }

   // output the child who says "S"
   while(cur != NULL)
   {
       int tmp = s;
       while(tmp-1 != 0)
       {
           cur = cur->next;
           tmp--;
       }
       
       cout<<cur->name<<"\n";
       if(cur->pre == cur && cur->next == cur) // the last element
           break;
       cur->pre->next = cur->next;
       cur->next->pre = cur->pre;
       Child* head = cur->next;
       delete cur;
       cur = head;
       
   }

    return 0;
}