10110 - Light, More Light

[Rajib Mazumder]

I can't understand why time limit exceed.... Can anybody help about the problem????


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


long primes[10000];

long pc;
long p1,p2;


int isprime(long num)
{
???????????????
}


void genprime(long lim)
{
???????????????????
}


int main()
{

long i,k,n;
genprime(32767);
while(cin>>n)
{
if(isprime(n)) cout<<"no";
else
{
for(i=2;i<=n-1;i++)
{
if((n%i)==0) k++;
}
if(k%2==0) cout<<"yes";
else cout<<"no";
}
cout<<endl;
}
return 0;
}

[Andrey Mokhov]

You needn't get all the divisors of the number - you just have to know whether it has even or odd number of divisors. There is a simple class of numbers that has odd number of divisors. Think of it. The problem can be solved in O(1) and you are trying to get AC in O(n). That won't work.

Good luck!

[laboni]

Here goes a hint - Square root the number , it has some special property, think of it.
bye

[fzrone]

well 2^32-1 means unsigned long
that should correct problem

[Calvin]

See my code :

Code: Select all

#include <stdio.h>

int main()

{
   unsigned int num,i,j;
   char step;

   scanf("%d",&num);

   while (num!=0)
   {
      step=0;
      i=4;
      j=5;

      while (num>i && i>0)
      {
         i+=j;
         j+=2;
      }

      if (num==i) printf("yes\n");
      else printf("no\n");
      scanf("%d",&num);
   }

   return 0;
}


I try it with the bigest number 2^32-1 and it's very fast, why does it take too much time ??

Please help me !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! i am driven to
despair !!

[Hisoka]

for this problem you unnecessary to solve by brute force, you can solve this with math. this is the logic:
1. the question only check the last lamp (ON/OFF).
2. you can know it with the factor from the number of last lamp.
3. if sum of factor is odd, the last lamp condition is ON, and when sum of
factor is even, the last lamp condition is OFF.
(I'm sorry for my bad english)
this is a example for you:
1. n=6.
the factor is 1 2 3 6.
the n factor is 4, and the result is OFF.
2. n=16.
the factor is 1 2 4 8 16.
the n factor is 5, and the result is ON.
from this you can solve it by math.

[Almost Human]

yes, that's right

but why I still got WA for this code :

Code: Select all

#include <math.h>
#include <stdio.h>

int main ( )
{
  unsigned long input , limit ;

/*  freopen ( "10110.in" , "r" , stdin ) ;
  freopen ( "10110.out" , "w" , stdout ) ;*/

  while ( 1 )
  {
    scanf ( "%li" , &input ) ;

    if ( input == 0 ) break ;

    limit = sqrt ( input ) ;

    if ( limit * limit == input )
      printf ( "yes\n" ) ;
    else
      printf ( "no\n" ) ;
  }

  return 0 ;
}


Explanation :
I use the square root of input to determine if the divisor of the input is odd or even. Am I right to use this algorythm ?

For Calvin :
try to use unsigned long instead of unsigned int

[deddy one]

yes the algorithm is right

int and long int is just the same here,
before I know that I also post something
like

Code: Select all

use long instead of int

what different here is long long int which
has 64 bit.

If I'm not mistaken int and long int
which are used here has 32 bit.

[Hisoka]

hello, almost human.......

for sqrt() your data type must use double, not integer.

[deddy one]

Hi Hisoka,

My Ac-ed program not using double,
I used long long int and get Acc

I don't know maybe just some
rounding precission.

[Almost Human]

I changed my code and use float instead of long and I 've got AC for this.

but I still wonder why I should changed it into float ?? Any suggestion ... ?

What I've changed :

Code: Select all

float input , limit ;


Code: Select all

scanf ( "%lf" , &input ) ;


Code: Select all

limit = ceil ( sqrt ( input ) ) ;


[Hisoka]

hello......

Deddy one can use long long for this problem. but I don't know about that, because as far as I know sqrt only work in float or another float. you can get more explanation from help in your bc 3.1.

Code: Select all

double sqrt(double);

[InOutMoTo]

I've got AC by using

[c]
double n, factor;

factor = floor( sqrt(n) );

if(n - factor*factor < 0.00000001)
printf("yes\n");
else
printf("no\n");[/c]

also got AC by using
[c]
long long n, factor;

factor = (long long) sqrt(n);

if(n == factor*factor)
printf("yes\n");
esle
printf("no\n");[/c]

But I think that the second method is more dangerous than the first one.
Because I force double turning into long long, possibly losing data.

For this prob, it's all ok

[Bug!]

Code: Select all

result=(long)sqrt(n);
  if(result*result==n)
     printf("yes\n");
  else printf("no\n");


is my algorithm correct??? I've tried with the sample input and it's work..
Please correct me if i'm wrong
Thanx..

[shamim]

Yes it is correct.