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): 88967    Accepted Submission(s): 33673

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

C语言程序代码

/*
题目大意：：
     先输入一个整数N，接着输入N行字符串，找出这N个字符串中出现次数最多的。 
解题思路：：
      先定义一个二维字符数组用来存放多行字符串，然后用两个for循环找出相同的
   字符串，并计数，然后找出最大值。并把此字符串输出。 
*/
#include<stdio.h>
#include<string.h>
int main(){
 int n,m,i,j,k,b[1001];
    char a[1001][20];
    while(scanf("%d",&n),n)
    {
     m=-1;
     getchar();
     for(i=0;i<n;i++)
      scanf("%s",&a[i]);
     for(i=0;i<n;i++)
     {
      for(j=0;j<i-1;j++)
      {
       if(strcmp(a[i],a[j])==0)
       {
        b[i]++;
       }
      }
     }
     for(i=0;i<n;i++)
     {
      if(m<b[i])
      {
       m=b[i];
       k=i;
      }
     }
     printf("%s\n",a[k]);
    }
    return 0;
}