Let the Balloon Rise—STL(map)

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 


这道题的题意就是给你几个不同颜色，输出颜色最多的那个颜色，要是所有颜色数量都相同，就输出第一个输入的颜色

这道题有很多种解法，一种就是用STL的map函数，也可以用字典树来做，我用了两天的时间用字典树来解决这个问题，但是一直有个bug，但是一直没有发现，然后我用map来做一下就AC了，啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊！！！！！！！


#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<map>
#include<string>
#include<iostream>
using namespace std;
int main()
{
    map<string,int> a;
    int i,j;
    string b;
    int n;
    int maxx;
    string d,ff;
    while(scanf("%d",&n)&&n)
    {
       a.clear();
       map<string,int>::iterator it;
       for(i=0;i<n;i++)
       {
           cin>>b;
           if(i==0)
            d=b;
           it=a.find(b);
           if(it==a.end())
           {
               a[b]=1;
           }
           else
           {
               a[b]++;
           }
       }
       maxx=-55;
       int flag=0;
       for(it=a.begin();it!=a.end();it++)
       {
           if(it->second!=a[d])
           {
               flag=1;
           }
           if(it->second>maxx)
           {
               maxx=it->second;
               ff=it->first;
           }
       }
        if(flag==0)
        {
            cout<<d<<endl;
        }
        else
        {
            cout<<ff<<endl;
        }
    }
}