Let the Balloon Rise

最新推荐文章于 2022-07-15 08:45:00 发布

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本文介绍了一个简单的编程问题：统计比赛中各种颜色气球的数量，并找出数量最多的那种颜色。通过使用C++中的map容器来记录每种颜色气球出现的次数，然后遍历map找到最多的颜色并输出。

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题目：

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：

输入一些不同颜色的气球，问哪个颜色的气球数量最多；

思路：

用map数组记录颜色所出现的次数；

使用前向迭代器遍历map，记录max值和数量最多的气球颜色；

最后输出气球颜色；

代码：

#include<stdio.h>
#include<iostream>
#include<string>
#include<map>
#include<algorithm>
using namespace std;

int main()
{
    int n;
    while(scanf("%d",&n)&&n)
    {
        string s1,s2;
        map<string,int>m;
        int i,maxx=-1;
        for(i=0;i<n;i++)
        {
            cin>>s1;
            m[s1]++;
        }
        map<string,int>::iterator it;//使用前向迭代器中序遍历map
        for(it=m.begin();it!=m.end();it++)
        {
            if((*it).second>maxx)
            {
                maxx=(*it).second;
                s2=(*it).first;
            }
        }
        cout<<s2<<endl;
    }
    return 0;
}