#include <stdio.h>
#include <string.h>
int main()
{
    char str[1001][1001]={0},str1[1001]={0},str2[1001]={0},empty[1001]={0};
    int i,j,N,k,n,h=0,p,flag;
        while(scanf("%d",&N)!=EOF&&N!=0)
        {
            k=0;
            h=0;
         for(i=0;i<N;i++)
        scanf("%s",str[i]);
        for(i=0;i<N;i++)
        {
        n=0;
            p=i;
            for(j=0;j<N;j++)
            {
                strcpy(str1,str[0]);
                if(strcmp(str[p],str[j])==0)
                {
                       n++;
                       strcpy(str1,str[j]);
                }
                k=n;
            

            }
            if(k>h)   
               strcpy(str2,str1);
            h=n;
        }
    printf("%s\n",str2);
       for(i=0;i<N;i++)
           for(j=0;j<1001;j++)
         for(i=0;i<1001;i++)
         {
            str1[i]=0;str2[i]=0;
         }
       }
    return 0;
}

请问题目要求是什么，要达到什么目的，LZ要写清楚啊

这个是杭电1004吧，我也纠结了很久
  
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7月16日管理员再次修正“虚拟比赛”的若干Bug，感谢热心用户的支持，期待大家继续测试~ 这里提建议
 
Let the Balloon Rise
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23450    Accepted Submission(s): 7100


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
 

没看出来，我OUT了，真正的高手也表现在查错上面

大家一起探讨探讨，谢谢楼上的关注

你程序的目的应该是判断并把重复的字符串输出吧。如果是的话，我建议你不要先先编码，先画个图出来看看
你程序的无用变量和循环结构太多。

我当时的AC代码：先看看，吃饭回来再看看楼主的代码……

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

typedef struct ballon
{
    char ch[16];
    int num;
};

int main()
{
    int n,point,i,max,t_p,flag;
    ballon ball[1001];
    char ch[16];
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
            break;
        point=1;
        getchar();
        for(i=0;i<n;i++)
        {
            ball[i].ch[0]='\0';
            ball[i].num=0;
        }
        while(n--)
        {
            scanf("%s",ch);
            for(flag=1,i=0;i<point;i++)
            {
                if(!strcmp(ch,ball[i].ch))
                {
                    ball[i].num++;
                    flag=0;
                    break;
                }
            }
            if(flag)
                strcpy(ball[point++].ch,ch);

        }
        for(max=i=0;i<point;i++)
        {
            if(ball[i].num>max)
            {
                max=ball[i].num;
                t_p=i;
            }
        }
        printf("%s\n",ball[t_p].ch);
    }
    return 0;
}

完了 毁了 彻底没了 这程序没救了 血压低 心电图是条直线 左胳膊和右胳膊挤在一堆了 救不了了

- -！看了下，还调试了下…没明白什么情况 ，楼主能注释一下再贴回来么……

我就是想用类似选择法的比较，用一个字符数组存储暂时个数最多的啊