Question

I am trying to find the sum of the multiples of 3 or 5 of all the numbers upto N.

This is a practise question on HackerEarth. I was able to pass all the test cases except 1. I get a time and memory exceeded error. I looked up the documentation and learnt that int can handle large numbers and the type bignum was removed.

I am still learning python and would appreciate any constructive feedback.

Could you please point me in the right direction so I can optimise the code myself?

test_cases = int(input())
for i in range(test_cases):
    user_input = int(input())
    sum = 0
    for j in range (0, user_input):
        if j % 3 == 0:
            sum = sum + j
        elif j % 5 == 0:
            sum = sum + j
    print(sum)

Answer 1

在这些问题中，尝试使用一些数学来找到一个直接的解决方案而不是强制它。

您可以在n下计算k的倍数，然后计算它们的总和。

例如，当k = 3且n = 13时，你有13 // 3 = 4倍他们的总和是3 * 1 + 3 * 2 + 3 * 3 + 3 * 4 = 3 *（1 + 2 + 3 + 4）

然后，使用关系：1 + 2 + .... + n = n *（n + 1）/ 2

要求3和5的倍数，你可以求和3的倍数，加上5的倍数之和，减去你计算两次的数：15的倍数。

所以，你可以这样做：

def sum_of_multiples_of(k, n):
    """
    Returns the sum of the multiples of k under n
    """
    # number of multiples of k between 1 and n
    m = n // k
    return k * m * (m+1) // 2

def sum_under(n):
    return (sum_of_multiples_of(3, n)
           + sum_of_multiples_of(5, n)
           - sum_of_multiples_of(15, n))

# 3+5+6+9+10 = 33
print(sum_under(10))
# 33

# 3+5+6+9+10+12+15+18 = 78
print(sum_under(19))
# 78

Time and memory limit exceeded - Python3 - Number Theory

1 个答案: