Given an Integer N. The task is to count numbers P less than N such that P is a product of two distinct perfect squares.
Examples:
Input : N = 36 Output : 5 Numbers are 4 = 12 * 22, 9 = 12 * 32, 16 = 12 * 42, 25 = 12 * 52, 36 = 12 * 62
Input : N = 1000000 Output : 999
Approach: Let us consider a number P = (a2 * b2) such that P <= N. So we have (a2 * b2) <= N. This can be written as (a * b) <= sqrt(N).
So we have to count pairs (a, b) such that (a * b) <= sqrt(N) and a <= b.
Let us take a number Q = (a * b) such that Q <= sqrt(N).
Taking a = 1, we have b = sqrt(N) – 1 numbers such that, ( a * b = Q <= sqrt(N)).
Thus we can have all sqrt(N) – 1 numbers such that (a2 * b2) <= N.
Below is the implementation of the above approach:
// C++ program to count number less // than N which are product of // any two perfect squares #include <bits/stdc++.h> using namespace std;
// Function to count number less // than N which are product of // any two perfect squares int countNumbers( int N)
{ return int ( sqrt (N)) - 1;
} // Driver program int main()
{ int N = 36;
cout << countNumbers(N);
return 0;
} |
// Java program to count number less // than N which are product of // any two perfect squares import java.util.*;
class solution
{ // Function to count number less // than N which are product of // any two perfect squares static int countNumbers( int N)
{ return ( int )Math.sqrt(N) - 1 ;
} // Driver program public static void main(String args[])
{ int N = 36 ;
System.out.println(countNumbers(N));
} } //This code is contributed by // Surendra_Gangwar |
# Python 3 program to count number # less than N which are product of # any two perfect squares import math
# Function to count number less # than N which are product of # any two perfect squares def countNumbers(N):
return int (math.sqrt(N)) - 1
# Driver Code if __name__ = = "__main__" :
N = 36
print (countNumbers(N))
# This code is contributed # by ChitraNayal |
// C# program to count number less // than N which are product of // any two perfect squares using System;
class GFG
{ // Function to count number less // than N which are product of // any two perfect squares static int countNumbers( int N)
{ return ( int )(Math.Sqrt(N)) - 1;
} // Driver Code public static void Main()
{ int N = 36;
Console.Write(countNumbers(N));
} } // This code is contributed // by Akanksha Rai |
<?php // PHP program to count number less // than N which are product of // any two perfect squares // Function to count number less // than N which are product of // any two perfect squares function countNumbers( $N )
{ return (int)(sqrt( $N )) - 1;
} // Driver Code $N = 36;
echo countNumbers( $N );
// This code is contributed by akt_mit ?> |
<script> // Javascript program to count number less
// than N which are product of
// any two perfect squares
// Function to count number less
// than N which are product of
// any two perfect squares
function countNumbers(N)
{
return parseInt(Math.sqrt(N), 10) - 1;
}
let N = 36;
document.write(countNumbers(N));
</script> |
5
Time Complexity: O(log(N)), Auxiliary Space: O(1)