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最新推荐文章于 2024-07-21 10:08:55 发布

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最新推荐文章于 2024-07-21 10:08:55 发布

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分类专栏： HDOJ

本文链接：https://blog.youkuaiyun.com/qq_14809723/article/details/63363594

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这是一个关于编程竞赛中统计最热门题目气球颜色的问题。输入包括多个测试案例，每个案例首先给出气球总数，随后列出每只气球的颜色。任务是找出出现次数最多的气球颜色。

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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): 117296 Accepted Submission(s): 45953

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

Author

WU, Jiazhi

Source

ZJCPC2004

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View Code

Problem : 1004 ( Let the Balloon Rise ) Judge Status : Accepted
RunId : 20161618 Language : C++ Author : 837274600
Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta

#include<stdio.h>
#include<string.h>
#include<math.h>

typedef struct Linknode
{
    char s[20];
    int n;
    struct Linknode *next;
}*Link;

int main()
{
    int n;
    while (scanf("%d", &n) != EOF&&n)
    {
        Link p, q, A = new Linknode;
        A->next = NULL;
        while (n--)
        {
            p = new Linknode;
            p->n = 0;
            p->next = NULL;
            scanf("%s", p->s);
            q = A;
            while (q)
            {
                if (!strcmp(p->s, q->s))
                {
                    q->n++;
                    q = q->next;
                }
                else if (q->next)
                    q = q->next;
                else
                    q->next = p;
            }
        }
        p = A;
        q = A->next;
        while (q)
        {
            if (q->n > p->n)
                p = q;
            q = q->next;
        }
        printf("%s\n", p->s);
    }
    return 0;
}