hdu 1004 Let the Balloon Rise

计赛气球统计

最新推荐文章于 2025-09-09 08:37:13 发布

转载最新推荐文章于 2025-09-09 08:37:13 发布 · 51 阅读

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原文链接：http://www.cnblogs.com/yazhou/p/3916800.html

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#java

本文介绍了一种通过编程解决竞赛中统计最受欢迎问题的方法，利用颜色代表不同题目，通过输入的颜色字符串来统计每种颜色（问题）出现的次数，并找出最多的一类。

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 74110 Accepted Submission(s): 27711

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

green

red

blue

red

pink

orange

pink

Sample Output

red

pink

字符串的比较：两个数组color[1005][16]和num[1005]。数组color是用来保存颜色的，数组num是用来保存颜色出现的次数，最后求出数组num中最大的数的下标即为出现次数最多的颜色。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 char color[1005][16];
 6 int num[1005];
 7 int main()
 8 {
 9     int n, i, j, max;
10     while(scanf("%d",&n) && n)
11     {
12         for (i=0; i<n; i++)
13             scanf("%s",&color[i]);
14         for(i=0; i<1005; i++)
15             num[i] = 0;
16         for(i=1; i<n; i++)
17             for(j=0; j<i; j++)
18                 if(strcmp(color[i],color[j])==0)
19                     num[i]++;
20         max = 0;
21         j = 0;
22         for(i=0; i<n; i++)
23             if (max<num[i])
24             {
25                 max = num[i];
26                 j = i;
27             }
28         printf("%s\n",color[j]);
29     }
30     return 0;
31 }