Python实现的求解最小公倍数算法示例

#!/usr/bin/python
from collections import Counter
def PrimeNum(num):
  r_value =[]
  for i in range(2,num+1):
   for j in range(2,i):
     if i % j == 0:
      break
   else:
     r_value.append(i)
  return r_value
def PrimeFactorSolve(num,prime_list):
  for n in prime_list:
   if num % n == 0:
     return [n,num / n]
def PrimeDivisor(num):
  num_temp =num
  prime_range= PrimeNum(num)
  ret_value =[]
  while num not in prime_range:
   factor_list= PrimeFactorSolve(num,prime_range)
   ret_value.append(factor_list[0])
   num =factor_list[1]
  else:
   ret_value.append(num)
  return Counter(ret_value)
def LeastCommonMultiple(num1,num2):
  dict1 =PrimeDivisor(num1)
  dict2 =PrimeDivisor(num2)
  least_common_multiple= 1
  for key in dict1:
   if key in dict2:
     if dict1[key] > dict2[key]:
      least_common_multiple*= (key ** dict1[key])
     else:
      least_common_multiple*= (key ** dict2[key])
  for key in dict1:
   if key not in dict2:
     least_common_multiple*= (key ** dict1[key])
  for key in dict2:
   if key not in dict1:
     least_common_multiple*= (key ** dict2[key])
  return least_common_multiple
print(LeastCommonMultiple(12,18))
print(LeastCommonMultiple(7,2))
print(LeastCommonMultiple(7,13))
print(LeastCommonMultiple(24,56))
print(LeastCommonMultiple(63,81))