Project Euler Problem 28
# n×nの一番外側の4個の数の和を返す def sum_of_group(n): last_num = n * n # 4個の数の最大値 diff = n - 1 # 4個の数の差 return (last_num + (last_num - diff) + (last_num - diff * 2) + (last_num - diff * 3)) answer = 1 for n in range(3, 1002, 2): answer += sum_of_group(n) print(answer)