Let the Balloon Rise

最新推荐文章于 2022-07-15 08:45:00 发布

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最新推荐文章于 2022-07-15 08:45:00 发布

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本文链接：https://blog.youkuaiyun.com/qq_43966523/article/details/84977162

本文介绍了一种算法，用于解决竞赛中预测最受欢迎问题的颜色。通过读取多种颜色字符串并统计每种颜色出现的频率，最终输出出现次数最多的颜色字符串。文章详细解释了算法流程，包括使用结构数组存储颜色和计数，以及比较和输出结果。

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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

问题链接：问题链接

问题简述：
读入多个颜色字符串；输出颜色出现次数最多的字符串。

问题分析：
建立结构数组，结构里放一个字符数组跟一个初始值为0的数字，每次输入后去前面输入过的颜色比较，若相等，前面的颜色所在的数字加一，最后输出改数字最大的颜色。

程序说明：
建立一个结构数组。里面放字符数组，跟一个为0的数字。

AC的C++程序如下：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
struct colour
{
char a[15];
int c;
}num[1001],p;
int main()
{
	char c[10];
	int n,g;
	while(cin>>n)
	{
		cin.ignore();
		g=n;
		for(int i=0;i<n;i++)
			{
				num[i].c=0;
				gets(num[i].a);
				for(int j=0;j<i;j++)
				{
					if	(strcmp(num[j].a,num[i].a)==0) num[j].c++;
				}
			}
			for(int i=0;i<n-1;i++)
				if(num[i].c>num[i+1].c)
				{
					p=num[i];
					num[i]=num[i+1];
					num[i+1]=p;
				}
			if(num[n-1].c>0) 
			{
				for(int i=0;num[n-1].a[i]!='\0';i++)
				{	cout<<num[n-1].a[i]; }
			cout<<endl;
			}	
			else
				for(int m=0;m<n;m++)
				{
					for(int i=0;num[m].a[i]!='\0';i++)
					cout<<num[m].a[i];
					cout<<endl;
				}
			if(g==0) break;	
		}
	return 0;
}