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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::marker::Chars;
use proconio::{fastout, input};

#[fastout]
fn main() {
    input! {_: usize, in_s: Chars};
    let mut rc = 0;
    let mut gc = 0;
    let mut bc = 0;
    let mut rcsum = Vec::new();
    let mut gcsum = Vec::new();
    let mut bcsum = Vec::new();
    let mut count = 0i64;

    for i in 0..in_s.len() {
        if in_s[i] == 'R' {
            rc += 1;
        } else if in_s[i] == 'G' {
            gc += 1;
        } else {
            bc += 1;
        }

        rcsum.push(rc);
        gcsum.push(gc);
        bcsum.push(bc);
    }

    for i in 0..in_s.len() {
        for j in i + 1..in_s.len() {
            if in_s[i] == in_s[j] {
                continue;
            }

            let c: char;

            if in_s[i] == 'R' {
                c = if in_s[j] == 'G' { 'B' } else { 'G' }
            } else if in_s[i] == 'G' {
                c = if in_s[j] == 'R' { 'B' } else { 'R' }
            } else {
                c = if in_s[j] == 'R' { 'G' } else { 'R' }
            }

            let mut v: i64;

            if c == 'R' {
                v = rcsum[rcsum.len() - 1] - rcsum[j];
            } else if c == 'G' {
                v = gcsum[rcsum.len() - 1] - gcsum[j];
            } else {
                v = bcsum[rcsum.len() - 1] - bcsum[j];
            }

            let k = j + (j - i);

            if k < in_s.len() && in_s[k] == c {
                v -= 1;
            }

            count += v;
        }
    }

    println!("{}", count);
}