Let the Balloon Rise

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

 
#include <iostream>

using namespace std;

int main()
{
    int num;
    while (cin>>num)//输入有多少个气球
    {
        string temp;
        int i,j;
        int count=0;//计数
        int max=0;//记录种类最多的气球的个数
        if(num==0) break;//当输入0个气球时结束
        string color[num];//气球的名称
        for(i=0;i<num;i++)//输入气球的颜色
        cin>>color[i];
        /*算法思想：从第一个气球开始与他自己和后面的气球比较,若相同
        则计数*/
        for(i=0;i<num;i++)
        {
            for(j=0;j<num;j++)
            {
            if(color[j]==color[i])
            count++;
            }
            if(count>max)//一直保留最大的那个数
            {
                max=count;
                temp=color[i];
            }

            count=0;

        }

       cout<<temp<<endl;

    }
    return 0;
}