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 "bits/stdc++.h"
using namespace std;

#define ASC(vec) vec.begin(), vec.end()    // 昇順ソート 例：sort(ASC(vec));
#define DESC(vec) vec.rbegin(), vec.rend() // 降順ソート 例:sort(DESC(vec));
#define rep(i, n) for (int i = 0; i < (n); i++)
#define Rep(i, n) for (int i = 1; i < n; i++)
#define REP(i, vec) for (auto i = vec.begin(); i != vec.end(); ++i)

const int       mod = 1000000007;
const int       inf = (1 << 21);
const long long INF = 1LL << 60;

using ii = pair<int, int>;
using ll = long long;
using vi = vector<int>;
using vd = vector<double>;
using vb = vector<bool>;
using vl = vector<ll>;
using vs = vector<string>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<ll>>;
using vvb = vector<vector<bool>>;
using vii = vector<pair<int, int>>;
using vll = vector<pair<ll, ll>>;

template<class T>
inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T>
inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
template<class T>
inline T GCD(T a, T b) {
    return b ? GCD(b, a % b) : a;
}
template<class T>
inline T LCM(T a, T b) {
    return (a + (b - 1)) / b;
}

int round_int(int a, int b) {
    return (a + (b - 1)) / b;
}

constexpr array<int, 9> dx = { 0, 1, 0, -1, -1, 1, 1, -1, 0 };
constexpr array<int, 9> dy = { 1, 0, -1, 0, 1, 1, -1, -1, 0 };

// ──────────────────────────────────────────────────────────────────

int dp[101][76];

bool isp(int a)
{
	for(int i=2;i*i<=a;i++)if(a%i==0)return 0;
	return 1;
}

int main() {
	int n;
	cin >> n;
	dp[0][1] = 1;
	int p = 0;
	for (int i = 2; i <= n; i++)
	{
		if (!isp(i))continue;
		int cnt = 0;
		for (int j = 1; j <= n; j++)
		{
			int a = j;
			while (a % i == 0)cnt++, a /= i;
		}
		for (int j = 0; j <= cnt; j++)
		{
			for (int k = 1; k * (j + 1) <= 75; k++)
			{
				dp[i][k * (j + 1)] += dp[p][k];
			}
		}
		p = i;
	}
	cout << dp[p][75] << endl;
}