10815 - Andy's First Dictionary

[sapnil]

To alamgir

Input this case:

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     aaa     a

>> your code given some garbage output


Thanks
Keep posting
Sapnil

[alamgir kabir]

I have got it Acc

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//code remove



Thank you very much sapnil.

[aeiou]

Hi ,
Deleted

[newton]

Why RE.
please any one check my small code

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#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define MAX 5001
using namespace std;

int comp(const void *a, const void *b)
{   
    return strcmp((char *)a,(char *)b);
}

int main(){
   char str[2000],*p;
   char word[MAX][100];
   char dil[] = " .:,\"'";
   int k = 0,i,j;
   //freopen("in.txt","rt",stdin);
   while(scanf("%s",str) == 1){
      p = strtok(str,dil);
      if(p)
         strcpy(word[k++],p);

      while(p){
         p = strtok(NULL,dil);
         if(p)         
            strcpy(word[k++],p);
      }      
   }
   for(i = 0; i<k; i++){
      for(j = 0;word[i][j]; j++){
         word[i][j] = tolower(word[i][j]);
      }
   }
   qsort(word,k,sizeof(word[0]),comp);
   for(i = 1; i < k+1; i++){
      if(strcmp(word[i],word[i-1]))
         printf("%s\n",word[i-1]);
   }
   return 0;
}

advanced thanks

[mf]

char word[MAX][100];

I think words are allowed to be up to 200 characters long by problem's text. And since it seems like you're using C++ anyway, why not just use std::string to keep strings?

Then, the problem text says "You can be sure that he number of distinct words in the text does not exceed 5000." But you're assuming that the *total* number of words is at most 5001, which is wrong.

char dil[] = " .:,\"'";

That's not going to solve the problem.

Stick to definitions, "a word is defined as a consecutive sequence of alphabets".

[Marcel Hernandez]

oh hai,

I've already solved the problem (scoring a pitiful run time btw) but I'm always not really sure of what to do when it comes to handle input like the one seen in this problem.


This is some kind of sketch written in C++ showing my approach:

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    string word;
    vector<string> dic;
    char symbol = getchar();
    while (symbol != EOF)
    {   
       if (isalpha(symbol)) // convert 'symbol' to lowercase if necessary and append it to the end of
                            // 'word'
       else // add 'word' to the vector and sort it if 'word' is not in it yet and finally clear the
            // auxiliary string
       symbol = getchar();       
    }


So I'd just like to know if there are better/easier/simpler ways to read several strings from a line. C code is also welcomed.
If only while (cin>> s) { ... } worked as if they were integers... -_-

[Jehad Uddin]

you can use getline or gets

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string str;
while(getline(cin,str,'\n'))
{
 .......then just take words ....
}
char str[300];
while(gets(str))
{
 .......
}


[sazzadcsedu]

Or you can use scanf in following way

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char S[20000][300];
char str[300];
char s[300];
while(scanf("%s",str)==1)
{
   k=0;
   word=0;
   for(i=0;str[i];i++)
   {
         
      if(isalpha(str[i]))
      {
         ch=tolower(str[i]);
         s[k]=ch;
         k++;
         word=1;
      }
      else
      {
         if(word==1)
         {
            s[k]='\0';
            strcpy(S[l],s);
            l++;
            word=0;
            k=0;
         }
      }

   }

   if(word==1)
   {
      s[k]='\0';
      strcpy(S[l],s);
      l++;
   }
}



[shantanu18]

I think the following approach is good to take input for this problem. It is difficult to include all punctuation.

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while((getline(cin,st))!=NULL)
{
   strcpy(ch,st.c_str());
   for(int i=0;i<(int)strlen(ch);i++)
   {
      if(!isalpha(ch[i]))
         ch[i]=' ';
      else if(isalpha(ch[i]) && isupper(ch[i])) ch[i]=tolower(ch[i]); //making all lower case
   }
   parse(ch);
}
//parse the char arr using strtok. there are only space char and tab char. Now it is easy to parse


[shaon_cse_cu08]

Plz Help me I m getting RE....

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

void ch_sort(char str[5000][201],int n);

int main()
{
   int n=0,i,j,tag=1;

   char a[5000][201],b[5000][201];

   while(gets(a[n++]))


   for(i=0;i<n;i++)
   {
       for(j=0;a[i][j]!='\0';j++)
      {
         if(a[i][j]>='A'&&a[i][j]<='Z')
            a[i][j]+=32;
      }
   }
   
   int k=0,l;

   for(i=0;i<n;i++)
   {
      l=0;
      tag=0;
       for(j=0;a[i][j]!='\0';j++)
      {
         if(a[i][j]>='a'&&a[i][j]<='z')
         {
            b[k][l++]=a[i][j];
            tag=1;
         }
         else
         {
            if(tag==1)
            {
               b[k][l]='\0';
               k++;
               l=0;
               tag=0;
            }
         }      
      }
      if(tag==1)
      {
         b[k][l]='\0';
         k++;
         l=0;
         tag=0;
      }

   }
   
   ch_sort(b,k);

   for(i=0;i<k;i++)
   {
      if(b[i][0]!='\0')
         puts(b[i]);
   }
      

return 0;
}

void ch_sort(char str[5000][201],int n)
{
   char dummy[201];
   
   int i,j;

   for(i=0;i<n-1;i++)
   {
      for(j=0;j<=n-i-1;j++)
      {
         if(strcmp(str[j],str[j+1])>0)
         {
         
            strcpy(dummy,str[j]);
            strcpy(str[j],str[j+1]);
            strcpy(str[j+1],dummy);
         }
         if(strcmp(str[j],str[j+1])==0)
            str[j+1][0]='\0';
      }
   }

return;
}

Plz plz Help....

[Victor Pangaldus]

Hi all,

I am newbie to Uva online judge. I am stuck on this problem (10815: Andy's First Dictionary) . My code give right outputs for the test case I get from the problem and in this forum but I continually get WA for this problem. I do not know what is wrong with my code. Can anyone help me to figure it out by giving me some critical test case?

I am looking forward to hearing from you. Thank you in advanced.

Regards,

Victor Pangaldus

[BUET]

My accepte program uses these delimiters:

" !\"#$%&'()*+,-./0123456789:;<=>?@[\\]^_`{|}~\t\r\n"

[kissu parina]

easy problem...just keep in mind that
andy's-->>
andy
s
not andys.....

[cool]

can any one help me please

i am getting TLE

[plamplam]

Use a faster sorting algorithm (I think you are using bubble-sort) to sort the words in dictionary order. I used qsort to get AC in 0.115 seconds. Good luck.