Find number of special numbers in a given range

Question

I am not able to reduce the complexity of this problem. Please give some better approach.

Is there a mathematical formula that i am not aware of or it can be done with better approach?

Description:

 A special number is not divisible by any number of the form Z*Z where (Z>1).

Problem: Find the number of special numbers in a given range.

Integer Limit:10^9

I have done it like this:

    import math
    def special(x):
        flag=1
        i=2
        if(x==0 or x==1 or x==2):
            return 1
        while(i*i <= x):               //This is the best i can think to limit the numbers.
            if(x%(i*i)==0):
                flag=0
                break
            i=i+1
        return flag

    t=int(raw_input())
    while(t):
        x,y=map(int,raw_input().split())
        count=0
        for i in xrange(x,y+1):
            if(special(i)):
                count+=1
        print(count)
        t=t-1

possible duplicate of [Count of squarefree numbers in range](http://stackoverflow.com/questions/23044332/count-of-squarefree-numbers-in-range) — Niklas B., Apr 15 '14 at 21:01

score 1 · Accepted Answer · edited May 23 '17 at 12:27

1

In special(x) you only have to iterate over primes less than or equal to sqrt(x). So I would precompute a list of primes (Fastest way to list all primes below N).

edited May 23 '17 at 12:27

Community

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answered Apr 15 '14 at 20:46

lukas1994

1,709
1
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Why only primes? Suppose number is 36 then its divisible by 6*6. So in the special function we need to check for 6 which is not prime. – Arvind Apr 15 '14 at 21:00

Find number of special numbers in a given range

1 Answers1