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Suppose we have two numbers P and Q and they form a number N = (P!/Q!). We have to reduce N to 1 by performing maximum number of operations possible. In each operation, one can replace N with N/X when N is divisible by X. We will return the maximum number of operations that can be possible.

So, if the input is like A = 7, B = 4, then the output will be 4 as N is 210 and the divisors are 2, 3, 5, 7.

To solve this, we will follow these steps −

N := 1000005

factors := an array of size N and fill with 0

From the main method, do the following −

for i in range 2 to N, do

if factors[i] is same as 0, then

for j in range i to N, update in each step by i, do

factors[j] := factors[j / i] + 1

for i in range 1 to N, do

factors[i] := factors[i] + factors[i - 1];

return factors[a] - factors[b]

Let us see the following implementation to get better understanding −

N = 1000005 factors = [0] * N; def get_prime_facts() : for i in range(2, N) : if (factors[i] == 0) : for j in range(i, N, i) : factors[j] = factors[j // i] + 1 for i in range(1, N) : factors[i] += factors[i - 1]; get_prime_facts(); a = 7; b = 4; print(factors[a] - factors[b])

7,4

4

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