Let the Balloon Rise

最新推荐文章于 2019-08-08 16:00:33 发布

CambridgeICPC

最新推荐文章于 2019-08-08 16:00:33 发布

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本文链接：https://blog.youkuaiyun.com/CambridgeICPC/article/details/81110586

本文介绍了一道经典的编程竞赛题目——气球统计问题。该问题要求从输入的颜色列表中找出出现次数最多的气球颜色。文章提供了一个使用C语言实现的解决方案，并详细解释了如何通过数组记录每种颜色的出现频率并找到最热门的颜色。

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Let the Balloon Rise

原题 HDU - 1004 链接 http://acm.hdu.edu.cn/showproblem.php?pid=1004

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

用number数组来记录相同气球的数量，每输入一个气球颜色，就遍历一遍，更新一遍number数值，用maxx来记录颜色数量最多的气球的下标，输出它的颜色即可。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

int main()
{
    int t;
    while(~scanf("%d",&t),t)
    {
        int number[10001];
        char book[10001][20];
        memset(number,0,sizeof number);
        for(int i=0; i<t; i++)
        {
            scanf("%s",book[i]);
            for(int j=0; j<i-1; j++)
            {
                if(strcmp(book[i],book[j])==0)
                {
                    number[i]++;
                }
            }
        }
        int maxx=0;
        for(int i=0; i<t; i++)
        {
            if(number[i]>=maxx)
            {
                maxx=i;
            }
        }
        printf("%s\n",book[maxx]);
    }
    return 0;
}