Suppose we are given two integer numbers m and a. Now n = p1(a + 1) *p2(a + 2) *...*pm(a + m), where pi is the i-th prime number and i > 0. We have to find the value of k, where k = summation of f(x) values of n. Here f(x) values are the number of divisor values of each divisor of n.
So, if the input is like m = 2, a = 1, then the output will be 60.
- So, n = 2^2 x 3^3
- n = 4 x 27
- n = 108
The divisors of 108 are: 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108
f(x) value of each divisor are: f(1) + f(2) + f(3) + f(4) + f(6) + f(9) + f(12) + f(18) + f(27) + f(36) + f(54) + f(108)
= 1 + 2 + 2 + 4 + 4 + 3 + 5 + 6 + 4 + 9 + 8 + 12
= 60.
To solve this, we will follow these steps −
- MOD := 10^9 + 7
- Define a function summ() . This will take n
- return floor value of ((n * (n + 1)) / 2)
- Define a function division() . This will take a, b, mod
- if a mod b is same as 0, then
- return floor value of a / b
- a := a + mod * division((-a modulo b), (mod modulo b), b)
- return floor value of (a / b) modulo mod
- if a mod b is same as 0, then
- mat := a new list containing a value 1
- while size of mat <= m + a, do
- insert (last element of mat * summ(len(mat)+1)) mod MOD at the end of mat
- return division(mat[m + a], mat[a], MOD)
Example
Let us see the following implementation to get better understanding −
MOD = 10**9 + 7 def summ(n): return ((n) * (n + 1)) // 2 def division(a, b, mod): if a % b == 0: return a // b a += mod * division((-a) % b, mod % b, b) return (a // b) % mod def solve(m, a): mat = [1] while len(mat) <= m + a: mat.append((mat[-1] * summ(len(mat)+1)) % MOD) return division(mat[m + a] , mat[a], MOD) print(solve(2, 1))
Input
2, 1
Output
60