ABC162 D - RGB Triplets

解答

code: python

n = int(input())

s = input()

ans = s.count('R') * s.count('G') * s.count('B')

for i in range(n):

for j in range(i + 1, n):

k = (j-i) + j

ans -= 1

print(ans)

メモ

提出

code: python

import itertools

n = int(input())

s = input()

# 等間隔ではない

# 2つ決めたら自動的に3つ目が決まる

# 計算量

# for i in itertools.combinations(range(4000), 2):

# print(i)

r = []

g = []

b = []

for i in range(len(s)):

r.append(i)

print(i)

g.append(i)

b.append(i)

setr = set(r)

setg = set(g)

setb = set(b)

print(r, g, b)

# どうやって等間隔, 順番異なるを引くか

print(len(r)*len(g)*len(b))

ans = len(r)*len(g)*len(b)

# R, G, Bの順番ってわけではない

# for i in r:

# for j in g:

# for k in b: