Problem:
Score: 400 points

Problem Statement

AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code.

The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code.

How many different PIN codes can he set this way?

Both the lucky number and the PIN code may begin with a 0.

Constraints

4 \leq N \leq 30000

S is a string of length N consisting of digits.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the number of different PIN codes Takahashi can set.

Sample Input 1

4
0224

Sample Output 1

3

Takahashi has the following options:

Erase the first digit of S and set 224.

Erase the second digit of S and set 024.

Erase the third digit of S and set 024.

Erase the fourth digit of S and set 022.

Thus, he can set three different PIN codes: 022, 024, and 224.

Sample Input 2

6
123123

Sample Output 2

17

Sample Input 3

19
3141592653589793238

Sample Output 3

329

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
#include <cmath>

using namespace std;

#define REP(i,n)	for(int i=0, i##_len=(n); i<i##_len; ++i)
#define REPS(i,s,n)	for(int i=(s), i##_len=(n); i<i##_len; ++i)

#define ALL(obj)	(obj).begin(),(obj).end()
#define sz(x) (int)(x).size()
#define v(T) vector<T>
#define VEC vector

typedef unsigned int 			uint;
typedef unsigned long long int	ull;
typedef long long int 			ll;

typedef pair<int,int>			P;
typedef vector<int>				Vint;
typedef vector<Vint>			VV;

template<typename T>string join(const v(T)&v)
{stringstream s;REP(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>inline ostream& operator<<(ostream&o,const v(T)&v)
{if(sz(v))o<<join(v);return o;}

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define MAX 		400010

const ll LINF = (ll)9e18+7;		//
const int INF = 100100100;			//2,147,483,647 (Max int)
const ll MOD = 1e9+7;
// テンプレ終了
//====================================================================//

// int DP[MAX];


//
int main(){
	int N;
	string S;
	cin >> N >> S;

	int ans = 0;
	map<string,int> used{};

	REP(i,N-2)REPS(j,i+1,N-1)REPS(k,j+1,N){
		string t = "aaa";
		t[0]=S[i],t[1]=S[j],t[2]=S[k];
		int tmp = used[t];
		//cout << t << " " << used[t] << endl;
		if(tmp>0) ;
		else {used[t] = 1;ans++;}
	}

	cout << ans << endl;

	return 0;
}