1004 Let the Balloon Rise

最新推荐文章于 2022-07-19 10:39:31 发布

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Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 118303 Accepted Submission(s): 46389

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

代码

#include <iostream>
#include <string>

using namespace std;

int main()
{
    int n;
    while(cin >> n && n)
    {
        string color[1001], temp;
        int count[1001] = {0};
        int max = 0, maxx = 0;
        for(int i = 0; i < n; i++)
        {
            cin >> temp;
            int j;
            for(j = 0; j < i; j++)
            {
                if(color[j] == temp)
                    ++count[j];
                if(count[j] > maxx)
                {
                    max = j;
                    maxx = count[j];
                }
            }
            if(i == j)
                color[i] = temp;
        }
        cout << color[max] << endl;
    }
    return 0;
}

讲真，这道题曾经困扰了我很长时间，就是因为自己想得太复杂了=.=

开一个string数组和int数组对应(貌似可以用map?)，及时更新最大值，最后输出string就可以了