E - Let the Balloon Rise HDU - 1004

最新推荐文章于 2021-01-15 15:09:31 发布

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本文介绍了一个简单的程序设计问题，即统计比赛中各颜色气球的数量来找出最受欢迎的问题。输入包括多个测试用例，每个用例首先给出气球总数，随后列出每只气球的颜色。输出则是出现次数最多的气球颜色。

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>

#include<cstdio>
#include<string>
#include<vector>
using namespace std;
int main()
{
    int n;

    while(scanf("%d",&n)&&n)
    {
      string s;
      vector<int>times;
      vector<string>color;
      for(int i=0;i<n;i++)
      {
          cin>>s;
          color.push_back(s);
      }
      for(int j=0;j<color.size();j++)
          times.push_back(1);
      for(int i=0;i<color.size();i++)
      {
           for(int j=i+1;j<color.size();j++)
         {
            if(color[i]==color[j])
            {
                times[j]++;
                times[i]++;
            }
         }
      }

      int max=0;
      string Smost;
      for(int i=0;i<times.size();i++)
      {
          if(times[i]>max)
          {
           max=times[i];
           Smost=color[i];
          }

      }
      cout<<Smost<<endl;

    }
    return 0;
}