Problem:
Score : 400 points

Problem Statement

We have a string S of length N consisting of R, G, and B.

Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:

S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.

j - i \neq k - j.

Constraints

1 \leq N \leq 4000

S is a string of length N consisting of R, G, and B.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the number of triplets in question.

Sample Input 1

4
RRGB

Sample Output 1

1

Only the triplet (1,~3,~4) satisfies both conditions. The triplet (2,~3,~4) satisfies the first condition but not the second, so it does not count.

Sample Input 2

39
RBRBGRBGGBBRRGBBRRRBGGBRBGBRBGBRBBBGBBB

Sample Output 2

1800

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORTASC(c) sort((c).begin(),(c).end())
#define SORTDESC(c,t) sort((c).begin(),(c).end(), std::greater<t>());
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define VMAX(a) *max_element(ALL(a))
#define VMIN(a) *min_element(ALL(a))
#define OUT(a) cout << (a) << endl;
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

const double PI  = acos(-1.0);
const long long MOD = 1000000007;

template <class T, class S>
using umap = std::unordered_map<T, S>;
using ll = long long;
using namespace std;

int main() {
    ll n, r = 0, g = 0, b = 0;
    string s;

    cin >> n >> s;

    REP(i, n) {
        if (s[i]=='R'){
            r++;
        } else if (s[i]=='G') {
            g++;
        } else {
            b++;
        }
    }

    ll cond1 = r*g*b;
    ll cond2 = 0;
    REP(i, ceil(n/2.0)-1){
        REP(j, n-2*(i+1)){
            if (s[j]!=s[j+i+1] && s[j+i+1]!=s[j+2*(i+1)] && s[j+2*(i+1)]!=s[j]){
                cond2++;
            }
        }
    }

  	cout << cond1-cond2 << endl;

    return 0;
}