杭电 1004[Let the Balloon Rise]

最新推荐文章于 2021-06-24 22:53:24 发布

原创最新推荐文章于 2021-06-24 22:53:24 发布 · 2.3k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#each #string #input #up

ACM 专栏收录该内容

13 篇文章

订阅专栏

本文分享了解决简单编程问题的策略和心得，包括输入解析、数据结构使用及优化思路，通过实例代码演示如何高效解决问题。

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

我居然被这种水题绊了大半天……好吧，我现在贴上我半天都不能AC的代码。

#include <iostream>
#include <string>
using namespace std;
void main()
{
    int n,i,j,max;
    while(cin>>n&&n!=0)
    {
        string str[1000], color;
        max=0;
        for(i=0;i<1000;i++) str[i]="";
        int num[1000]={0};
        for( i=1;i<=n;i++)
        {
            cin>>color;
            for( j=1;j<i;j++)
            {
                if(str[j].compare(color)==0 )
                {
                    num[j]++;    
                    break;
                }
            }
            if(j==i) 
            {
                str[i]=color;
                num[i]++;
            }
        }
        for(i=1;i<=n ;i++)
        {
            if(num[i]>num[max])
            {
                max=i;
            }
        }
        cout<<str[max]<<endl;
    }
}

好吧，只要把1000改成1001就可以AC了！

真是气得吐血哈哈。