hdu 1004 Let the Balloon Rise

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

5

green

red

blue

red

red

3

pink

orange

pink

0

 

 

Sample Output

red

pink

 

 

字符串的比較：兩個數組color[1005][16]和num[1005]。數組color是用來保存顏色的，數組num是用來保存顏色出現的次數，最後求出數組num中最大的數的下標即爲出現次數最多的顏色。

 

 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 }

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