HDOJ 1004 Let the Balloon Rise（水题）

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 88401    Accepted Submission(s): 33449

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



    
    
    
    在杭电做的前几道题，很水但是也是一个开始吧！
ac代码：

    
    
    
    #include<stdio.h> #include<string.h> int main() {     int n,i,j;     char str[1001][20];     char a[20];     int num[1001];     while(scanf("%d",&n)!=EOF&&n)     {         int k=0;         memset(num,0,sizeof(num));         for(i=0;i<n;i++)         {             scanf("%s",a);             int bz=0;             for(j=0;j<k;j++)             {                 if(strcmp(a,str[j])==0)                 {                     num[j]++;                     bz=1;                 }             }             if(bz==0)             {                 num[k]++;                 strcpy(str[k],a);                 k++;             }         }         int max=0,q;         for(i=0;i<k;i++)         {             if(num[i]>max)             {                 max=num[i];                 q=i;             }         }         printf("%s\n",str[q]);     }     return 0; }