关于map的练习 Easy - Let the Balloon Rise

最新推荐文章于 2025-11-25 00:06:26 发布

原创最新推荐文章于 2025-11-25 00:06:26 发布 · 94 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#算法

文章描述了一个编程问题，要求统计不同颜色的气球出现次数，输出出现次数最多的颜色。作者给出了初始和修改后的代码片段，强调了处理所有单词的重要性以避免越界错误。

我附上了我第一遍的错误代码第一遍的代码不能容纳所有单词而导致越界导致我WA了一次所以大家要注意这个问题

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

Inputcopy	Outputcopy
5 green red blue red red 3 pink orange pink 0	red pink

#include <bits/stdc++.h>
using namespace std;
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		map<string,int>mp;
		string a[3][1007];
		int n;
		cin >>n;
		for(int i=1;i<=3;i++)
		{
			for(int j=1;j<=n;j++)
			{
				cin >>a[i][j];
				mp[a[i][j]]++;
			}
		}
		for(int i=1;i<=3;i++)
		{
			int WOAIYUANSHEN=0;
			for(int j=1;j<=n;j++)
			{
				if(mp[a[i][j]]==1)WOAIYUANSHEN+=3;
				else if(mp[a[i][j]]==2)WOAIYUANSHEN+=1;
			}
			cout <<WOAIYUANSHEN<<" ";
		}
	}
}

下面是修改后的代码

#include <bits/stdc++.h>
using namespace std;
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		map<string,int>mp;
		string a[4][1007];
		int n;
		cin >>n;
		for(int i=1;i<=3;i++)
		{
			for(int j=1;j<=n;j++)
			{
				cin >>a[i][j];
				mp[a[i][j]]++;
			}
		}
		for(int i=1;i<=3;i++)
		{
			int WOAIYUANSHEN=0;
			for(int j=1;j<=n;j++)
			{
				if(mp[a[i][j]]==1)WOAIYUANSHEN+=3;
				else if(mp[a[i][j]]==2)WOAIYUANSHEN+=1;
			}
			cout <<WOAIYUANSHEN<<" ";
		}
	}
}