python - 给定总和的毕达哥拉斯三元组

Question

如果它等于输入，则以下代码打印毕达哥拉斯三元组，但问题是像 90,000 这样的大数需要很长时间才能回答。我可以做些什么来优化以下代码？1 ≤ n ≤ 90 000

def pythagoreanTriplet(n):

    # Considering triplets in
    # sorted order. The value
    # of first element in sorted
    # triplet can be at-most n/3.
    for i in range(1, int(n / 3) + 1):

        # The value of second element
        # must be less than equal to n/2
        for j in range(i + 1,
                       int(n / 2) + 1):

            k = n - i - j
            if (i * i + j * j == k * k):
                print(i, ", ", j, ", ",
                      k, sep="")
                return

    print("Impossible")
# Driver Code
vorodi = int(input())
pythagoreanTriplet(vorodi)

标签： pythonpythagoreantriplet

Answer 1

您的源代码对解决方案进行了蛮力搜索，因此速度很慢。

更快的代码

def solve_pythagorean_triplets(n):
  " Solves for triplets whose sum equals n "
  solutions = []
  for a in range(1, n):
    denom = 2*(n-a)
    num = 2*a**2 + n**2 - 2*n*a
    if denom > 0 and num % denom == 0:
      c = num // denom
      b = n - a - c
      if b > a:
        solutions.append((a, b, c))

  return solutions

操作码

修改了 OP 代码，因此它返回所有解决方案，而不是打印第一个找到的解决方案来比较性能

def pythagoreanTriplet(n): 

    # Considering triplets in  
    # sorted order. The value  
    # of first element in sorted  
    # triplet can be at-most n/3. 
    results = []
    for i in range(1, int(n / 3) + 1):  

        # The value of second element  
        # must be less than equal to n/2 
        for j in range(i + 1,  
                       int(n / 2) + 1):  

            k = n - i - j 
            if (i * i + j * j == k * k):
                results.append((i, j, k))

    return results

定时

 n     pythagoreanTriplet (OP Code)     solve_pythagorean_triplets (new)
  900   0.084 seconds                       0.039 seconds
  5000  3.130 seconds                       0.012 seconds
  90000 Timed out after several minutes     0.430 seconds

解释

函数solve_pythagorean_triplets是 O(n) 算法，其工作原理如下。

1. 正在寻找：

   a^2 + b^2 = c^2 (triplet)
   a + b + c = n   (sum equals input)
   

2. 通过搜索 a 来求解（即迭代的固定值）。有了固定值，我们有两个方程和两个未知数 (b, c)：

   b + c = n - a
   c^2 - b^2 = a^2
   

3. 解决方案是：

   denom = 2*(n-a)
   num = 2*a**2 + n**2 - 2*n*a
   if denom > 0 and num % denom == 0:
       c = num // denom
       b = n - a - c
       if b > a:
           (a, b, c) # is a solution
   

4. 迭代 range(1, n) 以获得不同的解决方案