10539 - Almost Prime Numbers

[Jan]

Your code doesnt even pass the samples. Check the input format.

[ishtiaq ahmed]

This problem is run time error signal eleven . Here is my code

Code: Select all

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

#define NUMBER 1000000
#define SIZE 1000000

int prime[ SIZE ] ;


void generate_prime( void )
{
	int i , j ,flag ; 
	prime[0] = 2  ; prime[1] = 3  ; prime[2] = 5  ; prime[3] = 7  ;
	prime[4] = 11 ; prime[5] = 13 ; prime[6] = 17 ; prime[7] = 19 ;
	int index = 7 ;
	for(i=21 ; i < NUMBER ; i += 2)
	{
		flag = 1 ;
		for(j = 0 ; i * prime[j] <= i ; j++)
		{
			if(i % prime[j] == 0)
			{
				flag = 0 ;
				break ;
			}
		}
		if( flag )
			prime[++ index] = i ;
	}

}


	

void generate_almost_prime( long long start , long long end )
{
	int i=0 , flag = 0 ;
	int sum_start , sum_end ;
	if( start != end)
	{
		sum_start = sum_end = 0;
		for(i= 0 ; prime[i] * prime[i] <= start ; i ++)
			sum_start += (int)( log10( start ) / log10( prime[i] ) ) - 1;
		for(i= 0 ; prime[i] * prime[i] <= end ; i ++)
			sum_end  += (int)( log10( end ) / log10( prime[i] ) ) -  1;
		printf("%d\n" , sum_end - sum_start ) ;
	}
	else
	{
		for(i= 0 ; prime[i] * prime[i] <= start ; i ++)
			if(start % prime[i])
			{
				flag++;
				if(flag == 2 )
					break ;
			}
		if(flag == 1 )
			printf("1\n");
		else
			printf("0\n");
	}

}


int main()
{
	int cas_no ,i ;
	long long start , end ;
	generate_prime() ;
	scanf("%d",&cas_no);
	for(i= 1 ; i <= cas_no ; i++)
	{
		scanf("%lld %lld",&start ,&end);
		generate_almost_prime( start , end );
	}
	return 0;
}

Plz try to help me.

[Shafaet_du]

You can solve it without binary search. so don't mess up your code with it. generate all the ALMOST PRIMES,store it in an array or vector,and iterate through it.

[zobayer]

for p and n, there are floor(log(n) / log(p)) powers of p from p to n. this formula can be used as well.
however, in implementation, log and log10 turn out to be too slow, even manual loop takes much much less time. also, one more thing is to note, although it doesn't have any effect on my machine, uva gives WA for log and AC for log10.
so be careful if you like "more mathematical accent" in this problem.

[plamplam]

Well well...it is always better to avoid floating point numbers. I got AC without the use of doubles and it runs fast. btw I agree with Shafet_DU. I got AC in 0.072 seconds without binary search although binary search gets me down to 0.024.

[human_being]

i understood the problem.
But in what type i took 10^12.
10^12 greater than the long long range

[aamurph]

Don't quite understand your question...

[raj]

Code: Select all

Accepted

[brianfry713]

http://en.wikipedia.org/wiki/Binary_search_algorithm

Java's long data type can handle well over 10^12
http://docs.oracle.com/javase/tutorial/ ... types.html

[mobarak.islam]

Here I'm getting wrong answer. please help

Code: Select all

Code removed after getting AC 

[brianfry713]

Try solving it without using floating point.

[mobarak.islam]

I solved it without floating point but still getting wrong answer

Code: Select all

Code removed after AC

[brianfry713]

sqrt and log return a double, try solving it without using them.

[mobarak.islam]

Got AC. Thanks brianfry713

[lighted]

I first calculate all the prime numbers between 1 and 10⁶
then for every prime numbers below low and high, I add all the log below and high to the base of the prime number as a and b. Then b-a is the answer.
But it is really inefficient, can anyone tell me some efficient algorithm.
I will be very grateful.

I got the same problem with you but now solved. Just don't use log.
Just calculate the lower bound and upper bound by brute force
I guess the log costs too much time.

for p and n, there are floor(log(n) / log(p)) powers of p from p to n. this formula can be used as well.
however, in implementation, log and log10 turn out to be too slow, even manual loop takes much much less time. also, one more thing is to note, although it doesn't have any effect on my machine, uva gives WA for log and AC for log10.
so be careful if you like "more mathematical accent" in this problem.

I used log and got accepted in 0.408.
So log can be used and it gives accepted.

I first calculate all the prime numbers between 1 and 10⁶
then for every prime numbers below low and high, I add all the log below and high to the base of the prime number as a and b. Then b-a is the answer.
But it is really inefficient, can anyone tell me some efficient algorithm.
I will be very grateful.

So i think it is efficient enough to get acc below 1second.

I found all primes in range 1..10^6 -> 78498 primes. For all of them i precalculated their log.