ZOJ Problem Set

Let the Balloon Rise

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

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Author: WU, Jiazhi
Source: Zhejiang Provincial Programming Contest 2004

题目大意：

输入n个气球颜色，输出颜色最多的

解题思路：

用STL模板中的map（key, value）来做很容易，key为颜色，当同一种颜色重复时，让value计数，然后将value值最大的那个key输出即可。

AC：

#include <cstdio>#include <string>#include <map>#include <iostream>using namespace std;int main(){int n;while(cin>>n && n){map <string, int> Balloon;string s;for(int i=0; i<n; i++){cin>>s;Balloon[s]++;}int iMax = 0;map <string, int>::iterator point, loc; //定义迭代器for(point=Balloon.begin(); point!=Balloon.end(); point++){if(iMax<point->second)//first指的是key，second指的是第二个value{iMax = point->second;loc = point;}}cout<<loc->first<<endl;}return 0;}