Author: Hitesh Ramchandani
Tester: Sagar Gupta
Editorialist: Hitesh Ramchandani
A good number is a number whose count of divisors is greater than the count of divisors for all the numbers less than it.
You have to find number of good numbers between two integers L and R(inclusive).
These good numbers are called highly composite numbers.
Let N = 2^a2 * 3^a3 * 5^a5 … p^ap (ie prime factorization of N ). A number N is a good number if :
- The primes 2, 3, 5 … p form consecutive primes in its factorization.
- The exponents are non-decreasing a2 ≥ a3 ≥ a5 … ≥ ap.
- The final exponent ap = 1, except for 2 cases when N = 4 (2^2) and N = 36 (2^2*3^2).
Naive approach(Subtask #1 30 pts):
Calculate all divisors of a number and store it which takes O(summation over i from 1 to n root(i)).
This is sufficient for 30 points and will fail for larger test cases.
Subtask #2(70 pts):
To answer the problem, we need to compute if a number N is good or not (or in other words highly composite). For this we need to calculate prime factorization of N. To calculate Prime factorization of N efficiently, we use Sieve-of-eratosthenes which calculates
prime numbers from 1 to N in O(Nlog^2(n)). So we first precompute all the prime numbers upto 10^7 using Sieve-of-eratosthenes.
Now for every N, we check if it satisfies the above conditions and also keep a count of number of divisors using the formula :
num of divisors = (a2+1)(a3+1)*(a4+1)…(ap+1).
Now consider N = 30 (2.3.5), it satisfies the above conditions but is still not a good number because N = 24 (2^3*3) also has 8 divisors. So it is necessary to store this information that number of divisors of 24 is 8 and thus even though 30 satisfies the above conditions, it is not a good number.
Precompute and store if N is good or not for all N’s from 1 to 10^7.
The number of good numbers upto 10^7 is just 48. One can hardcode those numbers and solve the problem.