方法3:使用functools.lru_cache(缓存减少重复计算) fromfunctoolsimportlru_cache@lru_cache(None)defsum_of_divisors(n):"""Compute sum of proper divisors using memoization."""ifn <=1:return0returnsum(i + (n // iifi != n // ielse0)foriinrange(1,int(n **0.5) +1)ifn % i ==0) - ...
defsum_of_proper_divisors(n):"""计算n的真因数之和"""total=1foriinrange(2,int(n**0.5)+1):ifn%i==0:total+=iifi!=n//i:total+=n//ireturntotal amicable_pairs=[]fornumberinrange(1,3000):sum_divisors=sum_of_proper_divisors(number)ifsum_divisors!=number:ifsum_of_proper_divisors(su...
return False # Find sum of their proper divisors sum_x = sum(e for e in range(1, x//2+1) if x % e == 0) sum_y = sum(e for e in range(1, y//2+1) if y % e == 0) #Return true of they satisfy the last condition return sum_x==y and sum_y==x print(are_amicab...
The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220. Evaluate the sum of all the amicable numbers under 10000. Answer: 31626 ''' def d(n): #计算数字n所有真因数之和 res = 0 for i in range(1, n//2+1): if n/i == float(n//i): res += i return(...
Write a Python program to calculate the sum of proper divisors for a given number and then determine if it is amicable with any other number in a specified range. Write a Python function that accepts a number and returns the sum of its amicable pair if it exists, or 0 if not. ...
14. Amicable Numbers Sum Write a Python program to find out if the given number is abundant. Note: Amicable numbers are two different numbers so related that the sum of the proper divisors of each is equal to the other number. (A proper divisor of a number is a positive factor of that...
A perfect number is one that is equal to the sum of its proper divisors, that is, sum of its positive divisors excluding the number itself. num = int(input("Enter any number: ")) sum = 0 for i in range(1, num): if(num % i == 0): sum = sum + i if (sum == num): pr...
A perfect number is a positive integer that is equal to the sum of its proper positive divisors. Problem statement Given alistof the integer numbers and we have to print all perfect numbers present in the given list. Printing perfect numbers from a list ...
("%d is divisor of %d\n",divisor,n); }}if (count == 0) printf("The number does not have proper divisors(it is prime)");else printf("%d and %d are the smallest and biggest divisors respectively", smallest, biggest);return 0;} 我没有使用两个for循环,而是将它们合并为一个,因为它们...
("%d is divisor of %d\n",divisor,n); }}if (count == 0) printf("The number does not have proper divisors(it is prime)");else printf("%d and %d are the smallest and biggest divisors respectively", smallest, biggest);return 0;} 我没有使用两个for循环,而是将它们合并为一个,因为它们...