Count numbers in a range having GCD of powers of prime factors equal to 1

Given a range represented by two positive integers L and R. The task is to count the numbers from the range having GCD of powers of prime factors equal to 1. In other words, if a number X has its prime factorization of the form 2p1 * 3p2 * 5p3 * … then the GCD of p1, p2, p3, … should be equal to 1.


Input: L = 2, R = 5
Output: 3
2, 3, and 5 are the required numbers having GCD of powers of prime factors equal to 1.
2 = 21
3 = 31
5 = 51

Input: L = 13, R = 20
Output: 7

Prerequisites: Perfect Powers in a Range
Naive Approach: Iterate over all numbers from L to R and prime factorise each number then calculate the GCD of powers of the prime factors. If the GCD = 1, increment a count variable and finally return it as the answer.

Efficient Approach: The key idea here is to notice that the valid numbers are not perfect powers since the powers of prime factors number are in such a way that their GCD is always greater than 1. In other words, all perfect powers are not valid numbers.
For e.g.

2500 is perfect power whose prime factorization is 2500 = 22 * 54. Now the GCD of (2, 4) = 2 which is greater than 1.
If some number has xth power of a factor in its prime factorization, then the powers of other prime factors will have to be multiples of x in order for the number to be invalid.

Hence, we can find the total number of perfect powers lying in the range and subtract it from the total numbers.

Below is the implementation of the above approach:





// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define N 1000005
#define MAX 1e18
// Vector to store powers greater than 3
vector<long int> powers;
// Set to store perfect squares
set<long int> squares;
// Set to store powers other than perfect squares
set<long int> s;
void powersPrecomputation()
    for (long int i = 2; i < N; i++) {
        // Pushing squares
        squares.insert(i * i);
        // if the values is already a perfect square means
        // present in the set
        if (squares.find(i) != squares.end())
        long int temp = i;
        // Run loop until some power of current number
        // doesn't exceed MAX
        while (i * i <= MAX / temp) {
            temp *= (i * i);
            // Pushing only odd powers as even power of a number 
            // can always be expressed as a perfect square 
            // which is already present in set squares
    // Inserting those sorted
    // values of set into a vector
    for (auto x : s)
long int calculateAnswer(long int L, long int R)
    // Precompute the powers
    // Calculate perfect squares in
    // range using sqrtl function
    long int perfectSquares = floor(sqrtl(R)) - floor(sqrtl(L - 1));
    // Calculate upper value of R
    // in vector using binary search
    long int high = upper_bound(powers.begin(),
                                powers.end(), R)
                    - powers.begin();
    // Calculate lower value of L
    // in vector using binary search
    long int low = lower_bound(powers.begin(),
                               powers.end(), L)
                   - powers.begin();
    // Calculate perfect powers
    long perfectPowers = perfectSquares + (high - low);
    // Compute final answer
    long ans = (R - L + 1) - perfectPowers;
    return ans;
// Driver Code
int main()
    long int L = 13, R = 20;
    cout << calculateAnswer(L, R);
    return 0;




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