HDOJ1004让气球飞吧

最新推荐文章于 2023-05-19 23:36:36 发布

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本文介绍了一个竞赛场景下统计最热门问题对应的气球颜色的程序设计方法。通过输入一系列颜色并统计每种颜色出现的次数，最终输出出现次数最多的颜色。

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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
思路：将输入的颜色存放到color缓存，与color_table比较，有相同的，则将该颜色的计数值加一，没有则存放到color_table表中，颜色输入结束，输出颜色
计数值最大的颜色。
#include<iostream>
#include<string>
using namespace std;
int main() {
	char color_table[1000][15]  //存放颜色   
    int count[1000], total;         
	char color[15];
	int i,j,flag,n, popular;
	while (cin>>n&&n!=0) 
	{
		total = 0;
		memset(count, 0, 1000);
		for (i = 0; i < n;i++) 
		{
			cin >> color;
			flag = 0;
			for (j = 0; j < i; j++)
			{
				if (strcmp(color_table[j], color) == 0)//如果输入的颜色与表中的某一颜色相同
                               {
					count[j]++;                    //该颜色个数+1
					flag = 1;
				}
			}
			if (flag == 0) {
				strcpy(color_table[total], color);
				count[i]++;
				total++;
			}
		}
		popular = 0;
		for (i = 0; i<total;i++){
			if (count[popular]< count[i]) {
				popular = i;
			}
		}
			cout << color_table[popular] << endl;
	}
	return 0;
}