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 Python (3.8.2) (judge verdict: Accepted)

def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])

    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])

    return arr

n=int(input())
alist=[0]*101
for i in range(2,n+1):
  for j in factorization(i):
    alist[j[0]]+=j[1]
if n==1:
  print(0)
else:
  ans=0
  for i in range(1,100):
    if alist[i]>=74:
      ans+=1
    if alist[i]>=24:
      for j in range(1,100):
        if i!=j and alist[j]>=2:
          ans+=1
    if alist[i]>=14:
      for j in range(1,100):
        if i!=j and alist[j]>=4:
          ans+=1
    if alist[i]>=2:
      for j in range(1,100):
        for k in range(j+1,100):
          if i!=j and  k!=i and alist[j]>=4 and alist[k]>=4:
            ans+=1
  print(ans)