Let the Balloon Rise 分类： HDU 2015-06-19 19:11 7人阅读 评论(0) 收藏

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 87647    Accepted Submission(s): 33130

Problem Description

Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result.

This year, they decide to leave this lovely job to you.

 

Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) -- the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters.

A test case with N = 0 terminates the input and this test case is not to be processed.

 

Output

For each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.

 

Sample Input

5
green
red
blue
red
red
3
pink
orange
pink
0

 

Sample Output

red
pink
想了好久，决定用链表了
#include <iostream>
using namespace std;
const int Max=1100;
struct node
{
    string s;
    int num;
    node *next;

};
int main()
{
    int n;
    string str;
    while(cin>>n,n)
    {
        node *head,*p,*q;
        head=new node;
        head->next=NULL;
        for(int i=0; i<n; i++)
        {
            cin>>str;
            q=head;
            p=head->next;
            while(p)
            {
                if(p->s==str)
                {
                    p->num++;
                    break;
                }
                p=p->next;
                q=q->next;
            }
            if(!p)
            {
                p=new node;
                p->s=str;
                p->num=1;
                p->next=NULL;
                q->next=p;
            }
        }
        p=head->next;
        q=head->next;
        while(p)
        {
            if(p->num>q->num)
            {
                q=p;
            }
            p=p->next;
        }
        cout<<q->s<<endl;

    }
    return 0;
}

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