895 - Word Problem

[rudy yulianto]

i have been try some input from Jan

abc
bcd
cde
def
eee
eff
eggg
ffff
#
e e e
#

and my output

1

but my submission is run time error

it's my code in C

Code: Select all

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

int counter[1232];
int main()
{
   int flag;
   char dic[1233][13];
   char puzzle[1234][13];
   int i =-1,j=-1;

   int a,b,len,c;
   char buff[13];
   int k;
   do
   {
      i++;
      gets(dic[i]);
   }while(dic[i][0]!='#');
   j=-1;
   do
   {
      j++;
      gets(puzzle[j]);
   }while(puzzle[j][0]!= '#');

   for(a=0;a<j;a++)
   {
      for(b=0;b<i;b++)
      {
         strcpy(buff,puzzle[a]);
         for(c=0;c<strlen(dic[b]);c++)
         {
            flag=1;
            for(k=0;k<strlen(buff);k++)
            {
               if(dic[b][c] != buff[k])
                  flag=0;

               else
               {
                  flag=1;
                  buff[k]=' ';
                  break;
               }
            }
            if(flag==0)break;
         }
         if(flag==1) counter[a]++;

      }
       printf("%d\n",counter[a]);
   }


   return 0;
}


please help me, give some reason why RTE and give me some IO..thx before...

[Jan]

How did you know that there are less than 1234 word puzzles? The problem states that the dictionary can have at most 1000 words. But there is no restriction for the number of word puzzles. So, just pick a puzzle, solve it and print the result until you get the '#'.

[Jehad Uddin]

Hello every body,i am getting frustrated with the problem.Though i passed all the I/Os from the board.Here is my code.Pls help me.

Code: Select all

#include<iostream>
#include<string>
using namespace std;
int main()
{
 char dic[1002][20];
 long int a[1002][27];
 long int word[27];
 long int i,j,k,l,m;
 char str[20];
 while(cin>>dic[0])
 {
  for(i=0;i<26;i++) a[0][i]=0;
    for(i=0;i<strlen(dic[0]);i++)
     a[0][dic[0][i]-97]++;
  k=1;
  while(1)
  {
   cin>>dic[k];
   if(strcmp(dic[k],"#")==0) break;
   for(i=0;i<26;i++) a[k][i]=0;
   for(i=0;i<strlen(dic[k]);i++)
     a[k][dic[k][i]-97]++;
   k++;
 
  }
  getchar();
  while(1)
  {
   gets(str);
   if(str[0]=='#') break;
   for(i=0;i<26;i++)
      word[i]=0;
   for(i=0;i<strlen(str);i++)
    if(str[i]>=97&&str[i]<=122)  word[str[i]-97]++;
   m=0;l=1;
   for(i=0;i<k;i++)
   {
    for(j=0,l=1;j<26;j++)
    if(word[j]<a[i][j])
   {
    l=0;
    break;
   }
   
    if(l) m++;
   }
 
  cout<<m<<endl;
 
  }
 
 
 
 }

 return 0;
}

Advanced thanks for the helpers.

[Shafaet_du]

sample

Code: Select all

hello
this
hiso
csedu
dhaka
niceprob
rivers
cricket
acm
acm
this
string
c
you
#
a c c c h i m
s r w d d g d
h s a i o c m
c a q r t r r
h l l y o o u
#

output:

Code: Select all

3
0
4
1
1

[receme]

Solved a moment ago ...I used LCS. It takes 0.020 sec.