673 - Parentheses Balance

[.pandre.]

[/quote]
Have you tried the special case of empty lines as input?
[/quote]

yes I have tried the empty string case and had the correct answer ("Yes")

And doesn't s.substring(i+2) give an exception when applied at the end of the string s?

no, no problem there


but I finnally got AC, i think the problem was that the number that I was passing to ReadLn() was too small, (in the accepted version, I used 50 for the number n (wow, they really use BIG numbers):
int n = Integer.parseInt(ReadLn(50).trim())

really stupid thing, don't you think?

anyway, thanks a lot for your time, i really apreciated your help

[Maniac]

I don't think the judge uses numbers of more than 25 digits for the number of test cases.. are you sure that this is the only change you made before getting AC? Well, congrats anyway and no problem, keep up the good work

[.pandre.]

I don't think the judge uses numbers of more than 25 digits for the number of test cases.. are you sure that this is the only change you made before getting AC? Well, congrats anyway and no problem, keep up the good work

yes, it was the only change ! quite weird, ins't it? well but that's an AC and that's what matters when we're so worried
once again thanks

[Maniac]

for your convenience and to avoid these problems in the future, there is also a readline function witch doesn't take a length-parameter but just reads a line of arbitrary length. See the forum-thread 'Java useful methods to save time'.

[faisneaz]

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

main()
{
char str[1000000],str1[1000000],c;
int a,b,size=0,counter=0,bool;
scanf("%d",&counter);
while(counter>0)
{
size=0;
bool=0;
scanf("%s",str);
//a=strlen(str);
if(strlen(str)==0)
{
printf("Yes\n");
}
else
{
for(b=0;str!='\0';++b)
{
if(str=='('||str=='[')
{
str1[size]=str;
str1[++size]='\0';
}
if(str==')'||str==']')
{
--size;
if((str==')'&&str1[size]!='(')||(str==']'&&str1[size]!='['))
{
if(strlen(str1)==0)
{
++bool;
}
break;
}
else
{
str1[size]='\0';
}
}
}
if(strlen(str1)!=0||bool>0)
{
printf("No\n");
bool=0;
}
else
{
printf("Yes\n");
}
}
--counter;
}
return 0;
}

[Karthekeyan]

Here's my code for the problem 673. I dunno why i am getting wrong answer again and again for this problem. Can someone help me?

Code: Select all

#include<iostream>
#include<stdio.h>
#include<vector>
using namespace std;

vector <char> st;
int check(char ch[200])
{
  int ans=1;
  int c=strlen(ch);
  for(int i=0;i<c;i++)
    {
      {
	if((ch[i]=='(')||(ch[i]=='['))
	  st.push_back(ch[i]);
	else
	  {
	    if(ch[i]==')')
	      {
		if(st.size()==0)
		  {
		    ans=0;
		    break;
		  }
		if(st.back()=='(')
		  st.pop_back();
		else
		  {
		    ans=0;
		    break;
		    }
		}
	      else
	      if(ch[i]==']')
		{
		  if(st.size()==0)
		    {
		      ans=0;
		      break;
		    }
		  if(st.back()=='[')
		    st.pop_back();
		  else
		    {
		      ans=0;
		      break;
		    }
		}
	    }
	}
    }
  if((st.size()!=0)||(ans==0))
    {
      cout<<"No\n";
      return 0;
    }
  else if((st.size()==0)&&(ans==1))
    {
      cout<<"Yes\n";
      return 1;
    }
}
main()
{
  int N,a;
  cin>>N;
  while(N>0)
    {
      st.clear();
      N--;
      char ch[200];
      cin>>ch;
      a=check(ch);
    }
}

thanx in advance

[emotional blind]

For this inputs

Code: Select all

3
((()

())))

your out put should be

Code: Select all

no
yes
no

but your program ignores the second case

[lolo]

i don't know why a got a WA
look my code and check it
please
thanks


#include <stdio.h>
#include <iostream.h>

bool EsI(char n)
{
if(n == '(' || n == '[')
return true;
return false;
}

char Contrario(char n)
{
if(n == '(')
return ')';
if(n == ')')
return '(';
if(n == '[')
return ']';
return '[';
}

int main()
{
char entrada[130] = {'\0'};
int cases , cant;

scanf("%d" , &cases);

while(cases--){

scanf("%s" ,entrada);

int contador = 0;

cant = strlen(entrada);


bool salto = false;


if(!cant){

printf("No\n");
salto = false;

}
else{

if(!(cant % 2)){

for (int i = 0 ; i < cant ; i ++){

if(EsI(entrada) && !salto){

char c = Contrario(entrada);
if(i == cant - 1){
salto = true;
}
for (int j = i + 1 ; j < cant && !salto ; j ++){

if(entrada[j] == entrada)
contador++;

if(entrada[j] == c && !contador--)
if(((j - i - 1) % 2))
salto = true;
}
contador = 0;
}
}
}
else
salto = true;
}
if(salto)
printf("No\n");
else
printf("Yes\n");
}
return 0;
}

[emotional blind]

Look at the problem description carefully

You are given a string consisting of parentheses () and []. A string of this type is said to be correct:

a)
if it is the empty string
(b)
if A and B are correct, AB is correct,
(c)
if A is correct, (A) and [A] is correct.

There may be empty string
here is a sample input

Code: Select all

3
([)]

((([[[]]])])

and output should be

Code: Select all

no
yes
no

your code ignors the empty string..

[murkho]

/*@BEGIN_OF_SOURCE_CODE*/

//No: 673...
//parenthess Balance.....


#include<stdio.h>

int main()
{
char str[150],*p,copy[150],ch;
int i,j,n,index = -1,flag;
fflush(stdin);
//freopen("c:\\673.txt","r",stdin);
scanf("%d",&n);
for(i = 1;i<=n;i++)
{
fflush(stdin);
gets(str);
flag = 0;
index = -1;
p = str;
while(*p)
{
if(*p =='(' || *p == '[')
{
ch = *p;
copy[++index] = ch; //pushing the character if it is '(' or '['.
}
if(*p ==')' || *p == ']')
{
if(index < 0 )
{
flag = 1;
break;
}

if( ( *p ==')' && copy[index] =='(') || (*p == ']' && copy[index] =='['))
{
index= index -1; //pop '(' or '[' when ')'or ']' is found..
}
else
{
flag = 1;
break;
}

}
p++;

}
if(index != -1)
flag =1;



if(flag ==1)
printf("No\n");
else
printf("Yes\n");


}


return 0;
}
/*END_OF_SOURCE_CODE*/

[L I M O N]

use STL stack

[valfiros]

Code: Select all

#include <stack>
#include <stdio.h>

using namespace std;
void sub();

void main()
{
	unsigned int case_num;
	scanf("%d",&case_num);
	for(unsigned int i=0;i<case_num;i++)
		sub();
}

void sub()
{
		stack <char> k; // stack
		unsigned int istrue=1; // flag
		char array[130];
		char * cpt;
		
		fflush(stdin);
		cpt = gets(array);

		while(*cpt != NULL)
		{
			if( (*cpt == '[') || (*cpt == '(') )
			{
				if(*cpt == '(')
					k.push(')');
				else
					k.push(']');
			}
			else if( (*cpt == ']') || (*cpt == ')') )
			{
				if(k.empty() == 1)
				{
					istrue=0;
					break;
				}
					
				if(*cpt == k.top())
					k.pop();
				else
				{
					istrue=0;
					break;
				}
			}
/*			else if( *cpt == ' ')
			{
				cpt++;
				continue;
*/			} // disabled!!!!
			else
			{
				istrue=0;
				break;
			}
			cpt++;
		}

		if (k.empty() == 0)
			istrue=0;
		if (istrue==1)
		{
			printf("Yes\n");
		}
		else
			printf("No\n");
}

i got wa... i cannot find any critical input. plz help!

[neno_uci]

Hi, your algorithm is correct, your problem is an i/o problem, as you are using gets, your first case string will be the end of the line(of the number of inputs line), so add an extra gets(anything) right after the scanf were you take the number of inputs and you will get AC...

[jaracz]

change scanf("%d",&case_num); to scanf("%d\n",&case_num); algo is fine
Regards

[komisarskip]

#include <stdio.h>
#include <string.h>
#define N 1000

int tab[N];
int k;

int main ()
{
int x;
char znak;
k=0;
scanf("%d\n",&x);
whhile: while (x--)
{
while((znak=getchar())!='\n' && znak!=EOF)
{
if(znak=='[') tab[k++]=1;
if(znak==']') if(k==0 || tab[--k]==2) {printf("No\n");while((znak=getchar())!='\n' && znak!=EOF);goto whhile;}
if(znak=='(') tab[k++]=2;
if(znak==')') if(k==0 || tab[--k]==1) {printf("No\n");while((znak=getchar())!='\n' && znak!=EOF);goto whhile;}
}
if(k!=0) printf("No\n");
else printf("Yes\n");
k=0;
}
return 0;
}