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 (3.4.3) (judge verdict: Accepted)

from collections import Counter

# 素因数分解
def prime_factorize(n):
    a = []
    while n % 2 == 0:
        a.append(2)
        n //= 2
    f = 3
    while f * f <= n:
        if n % f == 0:
            a.append(f)
            n //= f
        else:
            f += 2
    if n != 1:
        a.append(n)
    return a

n = int(input())
a = [0] * (n + 1)
for i in range(1, n + 1):
  c = prime_factorize(i)
  for i in c:
    a[i] += 1

a3, a5, a15, a25, a75 = 0, 0, 0, 0, 0
for i in a:
  if i >= 74:
    a75 += 1 
  if i >= 24:
    a25 += 1 
  if i >= 14:
    a15 += 1 
  if i >= 4:
    a5 += 1 
  if i >= 2:
    a3 += 1 

# 重複してか終えた分を減らす。a5は同じものを丸々ダブって数えるので２で割る。
ans = a75 + a25 * (a3 - 1) + a15 * (a5 - 1) + a5 * (a5 -1) * (a3 - 2) // 2 
print(ans)