Question

我已经解决了euler problem 12，但需要进行一些优化。我在euler论坛上看过，但它没有帮我优化它。但是，我已经设法获得解决方案，我只需要加快速度。它目前需要4分钟才能运行。这是我的代码：

import time

def nthtriangle(n):
    return (n * (n + 1)) / 2

def numberofnfactors(n):
    count = 0
    if n==1:
        return 1
    for i in range(1, 1 + int(n ** 0.5)):
        if n % i == 0:
            count += 2
    return count


def FirstTriangleNumberWithoverxdivisors(divisors):
    found = False
    counter = 1
    while not found:
        print(int(nthtriangle(counter)), "                   ", numberofnfactors(nthtriangle(counter)))
        if numberofnfactors(nthtriangle(counter)) > divisors:
            print(" first triangle with over ",divisors, " divisors is ", int(nthtriangle(counter)))
            found = True
            break
        counter += 1

start_time = time.time()
FirstTriangleNumberWithoverxdivisors(500)
print("My program took", time.time() - start_time, "to run")

Answer 1

不是单独计算每个三角形数字，而是使用生成器来获取三角形数字

from timeit import timeit

def triangle_numbers():
    count = 1
    num = 0
    while True:
        num += count
        count += 1
        yield num

def count_divisors(n):
    count = 0
    if n==1:
        return 1
    for i in range(1, 1 + int(n ** 0.5)):
        if n % i == 0:
            count += 2
    return count

print(timeit('next(num for num in triangle_numbers() if count_divisors(num) >= 500)', 
             globals=globals(), number=1))

在我的机器上给我3.8404819999996107（秒）。你也可以改进除数计数。

真正让你失望的是在你的循环中多次调用nthtriangle和numberoffactors！此外，对print的这些来电也是免费的。

欧拉12需要优化

1 个答案: