本题来自《程序员的算法趣题》中的第16题。
假设分别将3根长度相同的绳子摆成3个四边形。其中2根摆成长方形,另外1根摆成正方形。这时,当长度选择适当的话,会出现两个长方形的面积之和等于正方形的面积的情况(假设绳子长度和各四边形的边长都是整数)。
此外,将同比整数倍的结果看作是同一种解法。
令绳子长度为L=4*a,即正方形的边长为a。令另外两个长方形的较短的边分别为x1和x2,并且不失一般性可以假定它们满足关系:x1<=x2<a。
这个问题可以通过遍历搜索来解决。其中,对于任意的a,基于以上假设必然有,进一步可以推导出
。这样可以将x1的搜索范围缩小。代码如下:
- import sys
- import time
- import random
- from math import gcd, sqrt, ceil
- from typing import List
- # from queue import Queue
-
- class Solution:
-
- def lines2rectangles1(self, L:int)->int:
- """
- :L: The length of the lines
- :
- :ret: The number of the solutions
- """
- aMax = L // 4
-
- valid_cnt = 0
- for a in range(1,aMax+1):
- for x1 in range(1,ceil(a/sqrt(2))): # Assuming that x1 <= x2 < a, without loss of generality
- for x2 in range(x1,a):
- if x1*(2*a-x1) + x2*(2*a-x2) == a*a:
- if gcd(a,x1) == 1: # gcd(a,x1) --> gcd(a,x1,x2)
- valid_cnt = valid_cnt + 1
- # print('a = {0}, x1 = {1}, x2 = {2}'.format(a,x1,x2))
-
- return valid_cnt
其中,将同比整数倍的结果看作是同一种解法,这意味着,{a,x1,x2}必须满足三者互素,换句话说三者的最大公约数为1,才被计算为一个独立的答案。容易证明,在本题中,如果{a,x1,x2}满足题设要求的平方和关系的话,{a,x1,x2}三者互素等价于任何两者互素。因此,在代码中仅判断a和x1是否互素(用python中math模块中的gcd()函数)。
在解法1中,每次条件判断都是直接计算“x1*(2*a-x1) + x2*(2*a-x2) == a*a”,这会导致非常多的重复计算。事实上针对每个a(即针对最外层的每个循环),只需要计算一次a*a;同理针对每个x1(即针对第2层的每个循环),只需要计算一次x1*(2*a-x1)。这样可以将代码优化如下:
- def lines2rectangles2(self, L:int)->int:
- """
- :L: The length of the lines
- :
- :ret: The number of the solutions
- """
- aMax = L // 4
-
- valid_cnt = 0
- for a in range(1,aMax+1): # Assuming that x1 <= x2 < a, without loss of generality
- a_squ = a*a
- for x1 in range(1,ceil(a/sqrt(2))):
- area1 = x1*(2*a-x1)
- area2 = a_squ - area1
- for x2 in range(x1,a):
- if x2*(2*a-x2) == area2:
- if gcd(a,x1) == 1: # gcd(a,x1) --> gcd(a,x1,x2)
- valid_cnt = valid_cnt + 1
- # print('a = {0}, x1 = {1}, x2 = {2}'.format(a,x1,x2))
- return valid_cnt
运行结果表明这一简单的改进导致了运行时间下降了60%!
- def lines2rectangles3(self, L:int)->int:
- """
- :L: The length of the lines
- :
- :ret: The number of the solutions
- """
- aMax = L // 4
-
- valid_cnt = 0
- for a in range(1,aMax+1): # Assuming that x1 <= x2 < a, without loss of generality
- a_squ = a*a
- for x1 in range(1,ceil(a/sqrt(2))):
- if gcd(a,x1) == 1: # gcd(a,x1) --> gcd(a,x1,x2)
- area1 = x1*(2*a-x1)
- area2 = a_squ - area1
- for x2 in range(x1,a):
- if x2*(2*a-x2) == area2:
- valid_cnt = valid_cnt + 1
- # print('a = {0}, x1 = {1}, x2 = {2}'.format(a,x1,x2))
- return valid_cnt
运行结果表明这一简单的改进导致了运行时间进一步下降了30%!至少说明在本实现中,平方和关系的判断比互素关系的判断更为耗时。
进一步思考可以发现,当三个四边形满足题设条件时,假设其中一个长方形(不失一般性记为长方形1)的边长分别为a-x和a+x,则长方形2的面积必然为a*a – (a-x)(a+x)=x*x。反过来,令长方形2的边长分别为a-y和a+y,可以得到长方形1的面积必然为a*a – (a-y)(a+y)=y*y,也即三者的面积(的平方根)构成一组勾股数的关系!反之,容易证明,任何满足a*2=x*2+y*2的一组数{a,x,y}都对应着满足题设条件的三个四边形。由此可知,求满足题设条件下的解等价于寻找勾股数!
- def lines2rectangles4(self, L:int)->int:
- """
- :L: The length of the lines
- :
- :ret: The number of the solutions
- """
- aMax = L // 4
-
- valid_cnt = 0
- for a in range(1,aMax+1): # Assuming that x1 <= x2 < a, without loss of generality
- a_squ = a*a
- for x1 in range(1,ceil(a/sqrt(2))):
- if gcd(a,x1) == 1: # gcd(a,x1) --> gcd(a,x1,x2)
- diff = a_squ - x1*x1
- for x2 in range(ceil(a/sqrt(2)),a):
- if x2*x2 == diff:
- valid_cnt = valid_cnt + 1
- # print('a = {0}, x1 = {1}, x2 = {2}'.format(a,x1,x2))
- return valid_cnt
测试代码如下所示:
- if __name__ == '__main__':
-
- sln = Solution()
-
- L = 20
- tStart = time.time()
- num = sln.lines2rectangles1(L)
- tCost = time.time() - tStart
- print('L = {0}, numSlns = {1}, tCost = {2}(sec)'.format(L,num,tCost))
-
- L = 500
- tStart = time.time()
- num = sln.lines2rectangles1(L)
- tCost = time.time() - tStart
- print('#1: L = {0}, numSlns = {1}, tCost = {2}(sec)'.format(L,num,tCost))
- tStart = time.time()
- num = sln.lines2rectangles2(L)
- tCost = time.time() - tStart
- print('#2: L = {0}, numSlns = {1}, tCost = {2}(sec)'.format(L,num,tCost))
-
- L = 2000
- tStart = time.time()
- num = sln.lines2rectangles1(L)
- tCost = time.time() - tStart
- print('#1: L = {0}, numSlns = {1}, tCost = {2}(sec)'.format(L,num,tCost))
- tStart = time.time()
- num = sln.lines2rectangles2(L)
- tCost = time.time() - tStart
- print('#2: L = {0}, numSlns = {1}, tCost = {2}(sec)'.format(L,num,tCost))
- tStart = time.time()
- num = sln.lines2rectangles3(L)
- tCost = time.time() - tStart
- print('#3: L = {0}, numSlns = {1}, tCost = {2}(sec)'.format(L,num,tCost))
- tStart = time.time()
- num = sln.lines2rectangles4(L)
- tCost = time.time() - tStart
- print('#4: L = {0}, numSlns = {1}, tCost = {2}(sec)'.format(L,num,tCost))
运行后得到结果:
#1: L = 2000, numSlns = 80, tCost = 4.635588645935059(sec)
#2: L = 2000, numSlns = 80, tCost = 1.9797470569610596(sec)
#3: L = 2000, numSlns = 80, tCost = 1.2058100700378418(sec)
#4: L = 2000, numSlns = 80, tCost = 0.340120792388916(sec)
最终的算法所需要的时间只有初始算法的十分之一。