杭电ACM1004

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 107471    Accepted Submission(s): 41685

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

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

 int main()
 {
     int N=,max,e,i,j;
     char color[][];
     int count[]={};
     while(scanf("%d",&N)!=EOF&&N)
     {
         for(i=;i<N;i++)
             scanf("%s",color[i]);
         for(i=;i<N-;i++)
             {
                 for(j=i+;j<N;j++)
                 {
                     if(strcmp(color[i],color[j])==)
                         count[i]++;
                 }
             }
         max=count[];e=;
         for (i=;i<N;i++)
         {
             if (max<count[i]) e=i;
         }
         printf("%s\n",color[e]);
     }
     return ;
 }

关键点在于一开始就定义char color[1010][15],而不是char color[N][15]。毕竟不是C99标准，这样一开始就给出足够大的空间，反而更简单。