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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<fstream>
#include<sstream>
#include<algorithm>
#include<cstdio>
#include<cassert>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<string>
#include<deque>
#include<queue>
#include<vector>
#include<list>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<iomanip>
#include<utility>
#include<functional>
using namespace std;
int n,m;
int d[60][105];
int ans;
int prime[60];
int main()
{
	cin>>n;
	for(int i=2;i<=n;i++)
	{
		int t=i;
		for(int j=0;j<m;j++)
		{
			int p=prime[j];
			while(t%p==0)
			{
				t=t/p;
				d[j][i]++;
			}
		}
		if(t==i)
		{
			prime[m]=i;
			d[m][i]++;
			m++;
		}
	}
	for(int i=1;i<=n;i++)
		for(int j=0;j<m;j++)
			d[j][i]+=d[j][i-1];
	//1
	for(int j1=0;j1<m;j1++)
		if(d[j1][n]>=74)
			ans++;
	//2,3
	for(int j1=0;j1<m;j1++)
		for(int j2=0;j2<m;j2++)
		{
			if(j1!=j2&&d[j1][n]>=2&&d[j2][n]>=24)
				ans++;
			if(j1!=j2&&d[j1][n]>=4&&d[j2][n]>=14)
				ans++;
		}
	//4
	for(int j1=0;j1<m;j1++)
		for(int j2=0;j2<m;j2++)
			for(int j3=j2;j3<m;j3++)
				if(j1!=j2&&j2!=j3&&j1!=j3&&d[j1][n]>=2&&d[j2][n]>=4&&d[j3][n]>=4)
					ans++;
	cout<<ans<<endl;
	return 0;
}