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<string>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<utility>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<unordered_set>
#include<unordered_map>
#include<bitset>
using namespace std;
//#define MODE 1
#ifdef MODE
#define DEB(X) cout<< #X <<": "<<X<<" ";
#define ARDEB(i,X) cout<< #X <<"["<<i<<"]: "<<X[i]<<" ";
#define END cout<<endl;
#else
#define DEB(X) {}
#define ARDEB(i,X) {}
#define END {}
#endif
typedef long long ll;
#define int ll
#define uset unordered_set
#define umap unordered_map
typedef pair<int,int> P;
struct edge{int to,cost;};
#define REP(i,n) for(int i=0;i<(n);i++)
const int INF=100000000000000000;
const int MOD=1000000007;

std::vector<int> dec(int n){
	std::vector<int> v;
	int a=2;
	while(n>=a*a){
		if(n%a==0){
			v.push_back(a);
			n/=a;
		}
		else{
			a++;
		}
	}
	v.push_back(n);
	return v;
}

signed main(){
	int N;
	cin>>N;
	umap<int,int> prime;
	for(int i=1;i<=N;i++){
		vector<int> v=dec(i);
		REP(j,v.size()){
			prime[v[i]]++;
		}
	}
	int cnt[100]={};
	for(auto x:prime){
		if(x.second>=74){
			cnt[74]++;
		}
		else if(x.second>=24){
			cnt[24]++;
		}
		else if(x.second>=14){
			cnt[14]++;
		}
		else if(x.second>=4){
			cnt[4]++;
		}
		else if(x.second>=2){
			cnt[2]++;
		}
	}
	int ans=0;
	ans+=cnt[74];
	ans+=cnt[24]*(cnt[2]-1);
	ans+=cnt[14]*(cnt[4]-1);
	ans+=cnt[4]*(cnt[4]-1)/2*(cnt[2]-cnt[4]);
	ans+=(cnt[4]*(cnt[4]-1))/2*(cnt[4]-2);
	cout<<ans<<endl;
	return 0;
}