Let the Balloon Rise

最新推荐文章于 2021-02-27 19:36:55 发布

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#Let the Balloon Rise

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本文介绍了一种使用C++ map容器解决比赛中统计最热门气球颜色的方法。通过迭代器遍历统计每种颜色气球的数量，并找出出现次数最多的颜色。

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

简单来说，这是一个求哪个颜色出现的次数最多的问题。

从问题的描述来看。很适合用map容器来解决问题，正好再熟悉一下map容器的用法。

同时参考了一些有关于c++迭代器的用法

#include<cstdio>
#include<map>
#include<string>
#include<iostream>
#include<vector>
using namespace std;
map<string,int> a;
map<string, int>::iterator iter;  //设置一个名为iter的迭代器 
int main(){
	string b;	
	int n;
	while(cin>>n){
		if(n==0)
		break;
		string color="";
		a.clear();
		for(int i=0;i<n;i++){
			cin>>b;
			a[b]++;
		}
		int max=0;
		for(iter=a.begin();iter!=a.end();iter++){ //迭代器循环a的内容 
			if(iter->second>max){
				max=iter->second; //second代表键值 
				color=iter->first;//first代表键名 
			}
		}	
		cout<<color<<endl;	
	}	
}

附上c++迭代器的参考博客

C++迭代器博客