OJ Board

If I'm not wrong, there won't be such cases.

I suspect there are such cases.

Let me explain why.

The test for predicates is the most complicated so in my first attempt I simply tested for MOD, A, BA, DA, LA, and NAM then I assumed PREDA if it was none of the above. That is reasonable if all words have to fit in one of those classes. However I got a WA. When I redid my program to test explicitly for predicates and return an error if it came upon an invalid word I got AC. It seems to me that the judge's test cases include non-loglan words when the specs say it wouldn't.

Does your program have any problems with the following:

1) Extra spaces between words
2) Extra spaces at the END of the sentence (before the period)
3) Extra spaces at the START of the sentence
4) Extra letters AFTER the period
5) Extra spaces AFTER the period

etc.

junbin wrote:Does your program have any problems with the following:

1) Extra spaces between words
2) Extra spaces at the END of the sentence (before the period)
3) Extra spaces at the START of the sentence
4) Extra letters AFTER the period
5) Extra spaces AFTER the period

etc.

No 4. will trick my program. However, even when I fix that problem, I get a wrong answer if I assume that all words are valid. So I still think there are invalid words in the test cases. That would be ok, if the problem didn't say there would be no such words.

This always works:

[cpp]

bool predicate(std::string& word)
{
//returns true if word is a predicate
}

token get_next_token(const std::string& word)
{
if(word == ".")
return token(END, word);
else if(word=="a" || word=="e" || word=="i" || word=="o" || word=="u")
return token(A, word);
else if(word=="ga" || word=="ge" || word=="gi" || word=="go" || word=="gu")
return token(MOD, word);
else if(word=="ba" || word=="be" || word=="bi" || word=="bo" || word=="bu")
return token(BA, word);
else if(word=="da" || word=="de" || word=="di" || word=="do" || word=="du")
return token(DA, word);
else if(word=="la" || word=="le" || word=="li" || word=="lo" || word=="lu")
return token(LA, word);
else if(!vowel(word[word.length()-1]))
return token(NAM, word);
else if(predicate(word))
return token(PREDA, word);
else
return token(BAD, word);


}

[/cpp]



and this doesn't work:

[cpp]

token get_next_token(const std::string& word)
{
if(word == ".")
return token(END, word);
else if(word=="a" || word=="e" || word=="i" || word=="o" || word=="u")
return token(A, word);
else if(word=="ga" || word=="ge" || word=="gi" || word=="go" || word=="gu")
return token(MOD, word);
else if(word=="ba" || word=="be" || word=="bi" || word=="bo" || word=="bu")
return token(BA, word);
else if(word=="da" || word=="de" || word=="di" || word=="do" || word=="du")
return token(DA, word);
else if(word=="la" || word=="le" || word=="li" || word=="lo" || word=="lu")
return token(LA, word);
else if(!vowel(word[word.length()-1]))
return token(NAM, word);
else
return token(PREDA, word);


}

[/cpp]

What is the output of the following input set..

Input: Thanks in advance.

Jan, my AC program outputs the following for your input.

Output:
Bad!
Bad!
Bad!
Bad!

Hope this helps.

Thanks chunyi, But can you explain something...

The problem says..

Loglan words all end with a vowel; names, which are derived extra-linguistically, end with a consonant. Loglan words are divided into two classes--little words which specify the structure of a sentence, and predicates which have the form CCVCV or CVCCV where C represents a consonant and V represents a vowel

Therefore if we have a Loglan little word (which ends with a vowel but not a predicate) then do we skip it? or do we produce 'Bad!' ? .....

Suppose we have the following input

Input: Output: But the problem says...

Loglan words are divided into two classes--little words which specify the structure of a sentence

So why we are producing 'Bad!' for a little word?

Thank u again..

The problem description also states the subset of grammar considered. We are supposed to consider only the Loglan little words specified in the given grammar. No need to make things that complicated. So for the example you have given, the output will be:

Output:
Good
Bad!

I got Accepted

Thank u again to make the problem clear...

1. There is BAD formed words, words that are NOT in the grammar. This is very unfortunate because the problem statement says "you can assume all words will be correctly formed". I checked this by an assertion which catches bad words.

2. There is NO upper case characters. So you don't have to worry about case .

I am getting WA. here is my code:

#include<stdio.h>
#include<string.h>
int nword,type[1000];
char list[1000][100];

int VOW(char ch)
{
if(ch=='a' || ch=='e' || ch=='i' || ch=='o' || ch=='u')
return 1;
return 0;
}

int TAKE_PREDNAME(int s)
{
int i;
if(type[s]==7)
return s+1;

if(type[s]==5)
{
if(s==nword) return -1;

for(i=s+1;i<=nword;i++)
if(type[i]!=6)
break;

if(i==s+1) return -1;

return i;
}

return -1;
}

int TAKE_VERBPRED(int s)
{
int i;
if(type[s]!=2) return -1;

if(s==nword) return -1;

for(i=s+1;i<=nword;i++)
if(type[i]!=6)
break;

if(i==s+1) return -1;

return i;
}

int SENTENCE()
{
int k;
k=TAKE_PREDNAME(0);

if(k==-1 || k>nword) return 0;

k=TAKE_VERBPRED(k);

if(k==-1) return 0;
if(k>nword) return 1;

k=TAKE_PREDNAME(k);

if(k==-1 || k<=nword) return 0;
return 1;
}

int TAKE_PREDS(int s)
{
int i;
if(type[s]!=6) return -1;

for(i=s+1;i<=nword;i++)
if(type[i]!=1 && type[i]!=6)
break;

return i;
}

int PRODCLAIM()
{
int i,k;

TYPE1:
{
k=TAKE_PREDNAME(0);

if(k==-1) goto TYPE2;

if(type[k]!=3) goto TYPE2;
else k++;

if(k>nword) goto TYPE2;

k=TAKE_PREDS(k);

if(k==-1 || k<=nword) goto TYPE2;
else return 1;
}

TYPE2:
{
if(type[0]!=4) goto TYPE3;

k=TAKE_PREDS(1);

if(k==-1 || k<=nword) goto TYPE3;
else return 1;
}

TYPE3:
return 0;
}

int main()
{
int len,i,FALSE;
char ch,word[100];
freopen("test.in","r",stdin);
while(1)
{
len=-1;
nword=-1;
scanf("%c",&ch);
while( ch==' ')
scanf("%c",&ch);
while(ch!='#' && ch!='.')
{
if(ch=='\n')
{
scanf("%c",&ch);
continue;
}

if(ch==' ' && len>=0)
{
nword++;
word[len+1]='\0';
strcpy(list[nword],word);
scanf("%c",&ch);
len=-1;
continue;
}
if(ch==' ' && len<0)
{
scanf("%c",&ch);
continue;
}
len++;
if(ch>='A' && ch<='Z') ch=ch-'A'+'a';
word[len]=ch;
scanf("%c",&ch);
}

if(ch=='#')
break;

if(ch=='.' && len>=0)
{
nword++;
word[len+1]='\0';
strcpy(list[nword],word);
len=0;
}
FALSE=0;
/*if(ch=='.' && len==0)
{printf("Good\n"); continue;}*/
for(i=0;i<=nword;i++)
{
len=strlen(list[i]);
if(len==1 && VOW(list[i][0]))
type[i]=1;//A
else if(len==2 && list[i][0]=='g' && VOW(list[i][1]))
type[i]=2;//MOD
else if(len==2 && list[i][0]=='b' && VOW(list[i][1]))
type[i]=3;//BA
else if(len==2 && list[i][0]=='d' && VOW(list[i][1]))
type[i]=4;//DA
else if(len==2 && list[i][0]=='l' && VOW(list[i][1]))
type[i]=5;//LA
else if(len==5 && !VOW(list[i][0]) && !VOW(list[i][1]) && VOW(list[i][2]) && !VOW(list[i][3]) && VOW(list[i][4]))
type[i]=6;//PREDA
else if(len==5 && !VOW(list[i][0]) && VOW(list[i][1]) && !VOW(list[i][2]) && !VOW(list[i][3]) && VOW(list[i][4]))
type[i]=6;//PREDA
else
type[i]=7;//NAM
}

if((SENTENCE() || PRODCLAIM()))
printf("Good\n");
else
printf("Bad!\n");
}
return 0;
}

can any1 hlp me with debugging or can give me some cases??

(I have checked all the cases given in def posts!

Mahbub

I would just like to warn everyone who's trying to solve this problem that there are spaces after the period. Since this is not clearly stated in the input specification I think it should be mentioned here.

I would just like to warn everyone who's trying to solve this problem that there are spaces after the period. Since this is not clearly stated in the input specification I think it should be mentioned here.

I've got mant WAs in this problem, and I've viewed all the topics in this forum but still can't find the error.
Could anyone help me find the bug or give me some test cases?
Thanks in advance;

I did not read your code carefully....Where is your comments???

in any case is this your preds code If yes i think it is not valid
preds may be
predstring ||
predstring A predstring ||
predstring A predstring A predstring ||
predstring A predstring A predstring A predstring||

So it must end with predstring....

Also make sure in the end that you finish all strings