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): 65317    Accepted Submission(s): 24195

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

 

Source

ZJCPC2004  

//其实就是一个简单的算法题题，就是统计相同字符串出现的次数。然后输出出现次数最多的那个字符串。

首先可以把所有的字符串输入，然后用开头定义好的一个int型统计数组，用来统计每个字符串出现的次数。

但是我却不明不白wa了3次：

WA代码：

#include<stdio.h>
#include<string.h>
int main()
{
    int a[1050]={0},i,j,k,l,sum,n;
    char color[1010][20];
   while( ~scanf("%d",&n))
   {
        memset(a,0,sizeof(a));
       if(n)
       {
            for(i=1;i<=n;i++)
             scanf("%s",color[i]);
            for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
    {
             if(strlen(color[i])==strlen(color[j]))
        {
             for(k=0;k<strlen(color[i]);k++)
             if(color[i][k]!=color[i][k])
              break;
             if(k==strlen(color[i]))
              a[i]++;
        }
    }
          k=1;
          for(i=1;i<=n;i++)
          if(a[k]<a[i])
      k=i;
          printf("%s\n",color[k]);
       }
   }
return 0;
}

测试什么的都没有问题，可就是A不了。

如果大家有什么宝贵建议，请不吝请教

AC代码：

#include <stdio.h>
#include <string.h>
main(){
    int n, i, j, t, max, num[1000];
    char color[1000][16];
    while(scanf("%d", &n) != EOF){
        if(n){
            num[0]=0;
            scanf("%s", color[0]);
            for(i=1; i <n; i++){
                num[i]=0;
                scanf("%s", color[i]);
                for(j=0; j <i-1; j++)
                    if(strcmp(color[i], color[j])==0) num[i] +=1;
            }
            max=num[0];
            t=0;
            for(i=1; i <n; i++)
               if(max <num[i]) {max =num[i]; t=i;}
            printf("%s\n",color[t]);
        }
    }
}