2437: English word
Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 2362  Solved: 1137

Description
You still are worried about reading acm English problem, let me tell you a kind of very good method of Memorising Words, the root memory method,.the most interesting of the method is the prefix root. Now there are some English words to remembeand he knows some prefix root;He want to know how many words that can be remembered using each the prefix .in other words,how many words that contain the prefix ,Now to acmer you for help, can you help him?


Input
The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing a single integer number N that indicates the number of words (1 <= N <= 10000). Then exactly N lines follow, each containing a single word. Each word contains at least two and at most 10 lowercase characters, that means only letters 'a' through 'z' will appear in the word. The same word may appear several times in the list. Then follow a line containing a single integer number M that indicates the number of Words prefix question (0 <= M<= 10000). Then exactly M lines follow, each containing a single Words prefix.
Output
For the given input data, output exact M line, each line contain a ingle integer indicates the answer of each query.
Sample Input
1
4
preview
predict
premier
press
3
pre
press
pree

Sample Output
4
1
0

我的代码：
#include<stdio.h>
#include<string.h>
int main()
{
    int T,N,M,i,j,k,count,flag,e;
   char a[10000][10],b[10000][10];
    int c[10000];
    scanf("%d",&T);
    getchar();
    scanf("%d",&N);
    getchar();
    for(e=1;e<=T;e++)
    {
    for(i=0;i<=N-1;i++)
    {
        gets(a[i]);
    }
    scanf("%d",&M);
    getchar();
    for(i=0;i<=M-1;i++)
    {
        gets(b[i]);
        c[i]=strlen(b[i]);
    }
     for(i=0;i<=M-1;i++)//na b[1],b[2]...
     {
         count=0;
         for(k=0;k<=N-1;k++)
         {
          for(j=0;j<=c[i]-1;j++)
          {
              flag=0;
            if(b[i][j]!=a[k][j])
            {flag=1;break;}
           }
           if(flag==0)
           {count++;}
          }
       printf("%d\n",count);   
      }
  }
      return 0;
  }
   
跪谢啦~~

运行结果是对的。。。

楼主，你有好好读题目要求么

。。。是什么要求漏掉了？？

#include <stdio.h>
#include <stdlib.h>

struct node
{
    int flag;
    struct node* p[26];
}*head;

void find(char *s,struct node* head)
{
    int i,temp;
    (head->flag)++;

    if (s[0]==0) return;
    temp=s[0]-'a';
    if (head->p[temp]==NULL)
    {
        head->p[temp]=malloc(sizeof(struct node));
        head->p[temp]->flag=0;
        for (i=0; i<26; i++)
            head->p[temp]->p[i]=NULL;
    }
    find(s+1,head->p[temp]);
}

void output(char *s,struct node *head)
{
    while (head!=NULL && s[0]!=0) 
    {
        head=head->p[s[0]-'a'];
        s++;
    }
    if (head==NULL) printf("0\n"); 
        else printf("%d\n",head->flag);
}

int main()
{
    int T,N,M,i;
    char s[15];
    scanf("%d",&T);
    while (T--)
    {
        head=malloc(sizeof(struct node));
        head->flag=0; 
        for (i=0; i<26; i++) head->p[i]=NULL;

        scanf("%d",&N);
        while (N--)
        {
            scanf("%s",s);
            find(s,head);
        }

        scanf("%d",&M);
        while (M--)
        {
            scanf("%s",s);
            output(s,head);
        }
    }
}
        

收到的鲜花

• LeslieCh 于 2013-11-28 20:14 送鲜花  3朵   附言：好文章

谢谢你啊~成功AC了！