Question

我使用了强力方法并试图尽可能提高效率，但我只能在1小时后达到680的周长。我需要达到1000。

这是我试图解决的问题：

如果p是具有整数长度的直角三角形的周长>对于p = 120，恰好有三个解。{/，b}
{20,48,52}，{24,45,51}，{30,40,50}

对于p≤1000的值，最大化解的数量是多少？

在我的解决方案中，我为每一行添加了评论，以解释它的作用。

我使用过Python 3。

import math

# Initialisation of variables:
max = 0
ans = 0

# 'i' represents the perimeter. 'c' represents the hypotenuse length. 'a' and 'b' are both sides lengths:

# This outer loop iterates through all the perimeters from 1 to 1000.
for i in range(1, 1001):
    # Resets the count and arrA values.
    count = 0
    arrA = []
    # This loop iterates through all the hypotenuse values and makes sure its less than half of the perimeter:
    for c in range(1, i // 2):
        # This loop iterates through all the b values where b is lesser than (perimeter - hypotenuse):
        for b in range(1, i - c):
            # This loop iterates through all the a values where 'a' is lesser than (perimeter - hypotenuse - other side + 1):
            for a in range(1, i - c - b + 1):
                # Makes sure that all sides add up to the perimeter and that hypotenuse is lesser than the sum of the 2 other sides.
                if a + b + c != i or a + b <= c:
                    continue
                # Makes sure no instances are being repeated by and b being switched:
                if b in arrA:
                    break
                # Checks if its a right angled triangle:
                if math.sqrt(math.pow(a, 2) + math.pow(b, 2)) == c:
                    #Adds to the number of matches for the current perimeter.
                    count += 1
                    # Adds a to arrA so b can check later:
                    arrA.append(a)
                    # Extra output line for testing purposes (not needed):
                    print([f'{a:4}', f'{b:4}', f'{c:4}', f'{i:4}', f'{count:4}'])
    # checks if the current count was a new maximum or equal to previous maximum:
    if count >= max:
        max = count
        ans = i

#Prints final output:
print(f'Final Answer: {ans}')

以下是与上述完全相同的代码，但没有注释以便于阅读：

import math

max = 0
ans = 0

for i in range(1, 1001):
    count = 0
    arrA = []
    for c in range(1, i // 2):
        for b in range(1, i - c):
            for a in range(1, i - c - b + 1):
                if a + b + c != i or a + b <= c:
                    continue
                if b in arrA:
                    break
                if math.sqrt(math.pow(a, 2) + math.pow(b, 2)) == c:
                    count += 1
                    arrA.append(a)
                    print([f'{a:4}', f'{b:4}', f'{c:4}', f'{i:4}', f'{count:4}'])

    if count >= max:
        max = count
        ans = i

print(f'Final Answer: {ans}')

即使在2小时内也没有完成，但它的周长达到了800，所以我知道它工作正常，速度非常慢。

任何帮助都将不胜感激。

如何提高Project Euler＃39解决方案的效率？

0 个答案: