Project Euler Problem 28

Python

# 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)