Problem:
Score : 400 points

Problem Statement

You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?

Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.

Note

When a positive integer A divides a positive integer B, A is said to a divisor of B.
For example, 6 has four divisors: 1, 2, 3 and 6.

Constraints

1 \leq N \leq 100

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of the Shichi-Go numbers that are divisors of N!.

Sample Input 1

9

Sample Output 1

0

There are no Shichi-Go numbers among the divisors of 9! = 1 \times 2 \times ... \times 9 = 362880.

Sample Input 2

10

Sample Output 2

1

There is one Shichi-Go number among the divisors of 10! = 3628800: 32400.

Sample Input 3

100

Sample Output 3

543

Solution in PyPy3 (2.4.0) (judge verdict: Accepted)


def prime_factor(n):
    ass = []
    for i in range(2, int(n**0.5)+1):
        while n % i == 0:
            ass.append(i)
            n = n//i
    if n != 1:
        ass.append(n)
    return ass


n = int(input())
n_prime = {}
for i in range(1, n+1):
    for item in prime_factor(i):
        if item not in n_prime.keys():
            n_prime[item] = 0
        n_prime[item] += 1
#print(n_prime)
count_74 = 0
count_24 = 0
count_14 = 0
count_4 = 0
count_2 = 0
for prime, count in n_prime.items():
    if count >= 74:
        count_74 += 1
    if count >= 24:
        count_24 += 1
    if count >= 14:
        count_14 += 1
    if count >= 4:
        count_4 += 1
    if count >= 2:
        count_2 += 1
ans75 = count_74
ans25 = count_24*(count_2 - 1)
ans15 = count_14*(count_4 - 1)
ans5 = count_4*(count_4 - 1)*(count_2 - 2)//2
print(ans75 + ans25 + ans15 + ans5)