Count number of pairs (A <= N, B <= N) such that gcd (A , B) is B
Given a number n. We need find the number of ordered pairs of a and b such gcd(a, b) is b itself
Input : n = 2 Output : 3 (1, 1) (2, 2) and (2, 1) Input : n = 3 Output : 5 (1, 1) (2, 2) (3, 3) (2, 1) and (3, 1)
Naive approach : gcd(a, b) = b means b is a factor of a. So total number of pairs will be equal to sum of divisors for each a = 1 to n. Please refer find all divisors of a natural number for implementation.
Efficient approach : gcd(a, b) = b means that a is a multiple of b. So total number of pairs will be sum of number of multiples of each b (where b varies from 1 to n) which are less than or equal to n.
For a number i, number of multiples of i is less than or equal to floor(n/i). So what we need to do is just sum the floor(n/i) for each i = 1 to n and print it. But more optimizations can be done. floor(n/i) can have atmost 2*sqrt(n) values for i >= sqrt(n). floor(n/i) can vary from 1 to sqrt(n) and similarly for i = 1 to sqrt(n) floor(n/i) can have values from 1 to sqrt(n). So total of 2*sqrt(n) distinct values
let floor(n/i) = k k <= n/i < k + 1 n/k+1 < i <= n/k floor(n/k+1) < i <= floor(n/k) Thus for given k the largest value of i for which the floor(n/i) = k is floor(n/k) and all the set of i for which the floor(n/i) = k are consecutive
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