Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) 
Total Submission(s): 123205 Accepted Submission(s): 48540

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

解题思路：用map就很美滋滋,给map中插入时如果不成功，返回已有的节点给他加加，即给second(value)++

#include <map>
#include <iostream>
#include <string>

using namespace std;

int main()
{
    int num;
    while(scanf("%d",&num) == 1 && num){
        map<string,int> mp;
        int cnt = 0;
        string color;
        for (int i = 0; i < num; ++i)
        {
            string s;
            cin>>s;
            mp[s]++;
        }
        map<string,int>::iterator it = mp.begin();
        for (; it != mp.end(); ++it)
        {
            if (cnt < it->second)
            {
                cnt = it->second;
                color= it->first;
            }
        }
        cout<<color<<endl;
    }
    return 0;
}