题目链接：HDU 1004

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

题意

吹气球（2333）,输入很多不同颜色的气球，最后问什么颜色的气球最多，输出颜色

题解：

说实话，这道题我可能是走了很弯的路了，用了map来做，但是最后却没有用到map的优势，就当是练手速了，最后也是Wa了好几发才过。

建立气球类，表示这种气球的颜色，和当前数目，最后遍历一遍，更新最大数目的颜色，然后输出。

代码

#define _CRT_SECURE_NO_WARNINGS
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;

typedef long long ll;

const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 5;
struct Boll {
    string color;
    int ID;
    int Num = 0;
};
Boll B[2000];
int main() {
    int n;
    while (cin >> n, n) {
        //memset(B, 0, sizeof B);
        for (int i(1); i <= 2000; i++)
            B[i].Num = 0;
        map<string, int>P;
        string t;
        int M = 1;
        for (int i(0); i < n; i++) {
            cin >> t;

            if (!P[t]) {
                P[t] = M;
                B[M].color = t;
                B[M].ID = M;
                B[M].Num++;
                M++;
            }
            else {
                B[P[t]].Num++;
            }
        }
        string ans;
        long long MIN = -INF;
        for (int i(1); i <= n; i++) {
            if (MIN<B[i].Num) {
                MIN = B[i].Num;
                ans = B[i].color;
            }
        }
        cout << ans << endl;
    }

    return 0;
}