HDU 1004 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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本文将详细解读HDOJ平台上的编程挑战题HDOJ1004《让气球飘起来》，该题通过模拟气球比赛的情景，要求参赛者统计不同颜色气球的数量，从而找出最受欢迎的颜色。本文提供了输入输出示例，并给出了一段C++代码实现解决方案。

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http://acm.hdu.edu.cn/showproblem.php?pid=1004

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 90418 Accepted Submission(s): 34348

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

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
    int n,i,j,max,r,k,sum;
    char a[1001][16];
    while(cin>> n && n)
    {
        getchar();
        for(i=0;i<n;i++) gets(a[i]);
        max=1;
        k=0;
        for(i=0;i<n;i++)
        {
            sum=0;
            if(a[i][0]== '\0') continue;
            for(j=i+1;j<n;j++)
            {
                if(strcmp(a[i],a[j])==0)
                   {
                       a[j][0]= '\0';
                       sum++;
                   }
            }
            if(sum>max)
            {
                max=sum;
                k=i;
            }
        }
        puts(a[k]);
    }
    return 0;
}