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: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <cmath>
#include <complex>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <bitset>
#include <algorithm>
#include <numeric>
#include <functional>
#include <cassert>
#include <iomanip>
using namespace std;

#define Rep(b, e, i) for(int i = b; i <= e; i++)
#define Repr(e, b, i) for(int i = e; i >= b; i--)
#define rep(n, i) Rep(0, n-1, i)
#define repr(n, i) Repr(n-1, 0, i)
#define all(v) (v).begin(), (v).end()
#define pb(v) push_back(v)
#define uniq(v) (v).erase(unique(all(v)),(v).end())
#define bitcnt(x) __builtin_popcount(x)
#define fst first
#define snd second
#define Pqaz(T) priority_queue<T,vector<T>,greater<T>>
#define Pqza(T) priority_queue<T>
#define put(x) cout << x;
#define putsp(x) cout << x << ' ';
#define putbd cout << "---------------------------------------------" << endl;
#define putln(x) cout << x << endl;
#define debug(x) cerr << #x << "=" << x << endl;
#define ENJYU std::ios::sync_with_stdio(false);std::cin.tie(0);

typedef long long ll;
typedef pair<ll, ll> llP;
typedef pair<int, int> intP;
typedef complex<double> comp;
typedef vector <int> vec;
typedef vector <ll> vecll;
typedef vector <double> vecd;
typedef vector <vec> mat;
typedef vector <vecll> matll;
typedef vector <vecd> matd;

//vector の中身を出力
template <class T>ostream &operator<<(ostream &o,const vector<T>&v)
{o<<"{";for(int i=0;i<(int)v.size();i++)o<<(i>0?", ":"")<<v[i];o<<"}";return o;}

const int MAX = 200020;
const double PI = 3.14159265358979323846;
const double EPS = 1e-12;
const int INF = 1<<29;
const ll INFL = 1e18;
const ll MOD = 1000000007;

const int dx[]={1,0,-1,0},dy[]={0,1,0,-1};
//const int dx[]={-1,-1,-1,0,1,1,1,0},dy[]={-1,0,1,1,1,0,-1,-1};

//********************************template END****************************************//

vector<int> enum_prime(int N)
{
	bool prime[N];
	memset(prime, true, sizeof(prime));
	prime[0] = prime[1] = false;

	for (int i = 2; i < N; i++)
	{
		if (!prime[i])
			continue;
		for (int j = 2; i * j < N; j++)
		{
			prime[i * j] = false;
		}
	}

	vector<int> ps;
	rep(N, i)
	{
		if (prime[i])
			ps.pb(i);
	}

	return ps;
}

void solve(void)
{
	int N;
	cin >> N;

	vec ps = enum_prime(N);

	cout << ps << endl;

	map<int, int> mp, cnt;
	vec as = vec{3,5,15,25,75};

	Rep(2, N, n)
	{
		int x = n;
		for (int p : ps)
		{
			while (x % p == 0)
			{
				mp[p]++;
				x /= p;
			}
		}
	}

	for (auto p : mp) for (int a : as)
	{
		if (a-1 <= p.snd)
		{
			cnt[a]++;
		}
	}

	//for (auto p : cnt){cout << p.fst << ' ' << p.snd << endl;}

	int ans = max(0, (cnt[3]-2) * cnt[5] * (cnt[5]-1) / 2);
	ans += max(0, cnt[75]);
	ans += max(0, (cnt[5]-1) * cnt[15]);
	ans += max(0, (cnt[3]-1) * cnt[25]);

	cout << ans << endl;
}

int main(void)
{
	solve();
	//cout << "yui(*-v・)yui" << endl;
	return 0;
}