ACM Let the Balloon Rise

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.

InputInput 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. 
OutputFor 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

代码逻辑值得玩味

 /*
     Name: Let the Balloon Rise
     Copyright:
     Author:
     Date: 09/08/17 17:01
     Description: 寻找颜色出现次数最多的气球
 */
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 using namespace std;
 int main()
 {
     int n;
     int num[];
     char ch[][];
     while(cin>>n && n)
     {
         memset(num,,sizeof(num));
         for(int i = ; i < n; i++)
         {
             cin>>ch[i];
             for(int j = ; j < i; j++)
                 if(strcmp(ch[j],ch[i]) == ) num[i]++;
         }
         int max = num[];
         int t = ;
         for(int i = ; i < n; i++)
         {
             if(max < num[i])
             {
                 max = num[i];
                 t = i;
             }
          }
          cout<<ch[t]<<endl;
     }

     return ;
 }