Let the Balloon Rise（Map容器）

喜欢ヅ旅行 2022-08-05 05:23 99阅读 0赞

# Let the Balloon Rise #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 87369    Accepted Submission(s): 33024**

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

Author

WU, Jiazhi

#include<cstdio>
    #include<iostream>
    #include<map>//引入map容器
    using namespace std;
    int main()
    {
        int n,num;
        string s,ans;
        while(cin>>n&&n)
        {
            num=-5;//存数字
            map<string ,int >word;//声明一个关键字为字符串值为数字的map
            map<string ,int>::iterator it;
            while(n--)
            {
                cin>>s;
                word[s]++;//关键字的值加一
            }
            for(it=word.begin();it!=word.end();it++)
            {
                if(it->second>num)
                {
                    num=it->second;
                    ans=it->first;
                }
            }
            cout<<ans<<endl;
        }
    }