map- Let the Balloon Rise

本文介绍了一种通过使用C++中的map数据结构来统计比赛中各颜色气球出现次数的方法,并给出了完整的代码实现。该程序能够找出每轮比赛中最受欢迎的问题所对应的气球颜色。

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map- Let the Balloon Rise

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<string>
#include<cstdio>
#include<iostream>
#include<map>
using namespace std;

map<string,int> m;
int main(){
	int n;
	while(cin>>n&&n!=0)
	{
		int num=0;
		for(int i=0;i<n;i++){
			string a;
			cin>>a;
			m[a]++;
		} 
		string b;
		for(map<string,int>::iterator ite=m.begin();ite!=m.end();ite++)
		{
			if(num<ite->second){
				num=ite->second;
				b=ite->first;
			}
		}
		cout<<b<<endl;
		m.clear();
	}
	return 0;
}

另一种

#include<iostream>
#include<cstdio>
#include<string>
#include<map>
using namespace std;
map<string,int> m;

string a;
int main()
{
	int n;
	while(1)
	{/*做无限循环直到为零跳出循环*/ 
		cin>>n;
		if(n==0)
		{
			break;
		}	
		m.clear();/*清空map里的键值*/
		for(int i=0;i<n;i++)
		{/*将字符串依次存入a*/
			cin>>a;
				if(0==m.count(a))
			{
				m.insert(make_pair(a,1));
			} 
			else 
			{
				m[a]++;
			}/*如果map里没有相映的字符串就设定一个键并给值1,有就加一*/
		}
		map<string,int>::iterator ite;	/*迭代器用来查询*/
		pair<string,int> pa=make_pair("\0",-1);
		/*定义用来存入当前最大的数据*/
		for(ite=m.begin();ite!=m.end();ite++)
		{/*依次查询map的值*/
			ite->second;
			if(ite->second>pa.second)
			{/*用来存入当前最大的数据*/
				pa=*ite;
			}
		}/*pa.first存的是字符串,pa.second存的是字符串的个数*/
		cout<<pa.first<<endl;
	}
	return 0;
}
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