Project Euler：Problem 88 Product-sum numbers

A natural number, N, that can be written as the sum and product of a given set of at least two natural numbers, {a1, a2, ... , ak}
is called a product-sum number: N = a1 + a2 + ... + ak = a1 × a2 × ... × ak.

For example, 6 = 1 + 2 + 3 = 1 × 2 × 3.

For a given set of size, k, we shall call the smallest N with this property a minimal product-sum number. The minimal product-sum numbers for sets of size, k =
2, 3, 4, 5, and 6 are as follows.

k=2: 4 = 2 × 2 = 2 + 2

k=3: 6 = 1 × 2 × 3 = 1 + 2 + 3

k=4: 8 = 1 × 1 × 2 × 4 = 1 + 1 + 2 + 4

k=5: 8 = 1 × 1 × 2 × 2 × 2 = 1 + 1 + 2 + 2 + 2

k=6: 12 = 1 × 1 × 1 × 1 × 2 × 6 = 1 + 1 + 1 + 1 + 2 + 6

Hence for 2≤k≤6, the sum of all the minimal product-sum numbers is 4+6+8+12 = 30; note that 8 is only counted once in the sum.

In fact, as the complete set of minimal product-sum numbers for 2≤k≤12 is {4, 6, 8, 12, 15, 16}, the sum is 61.

What is the sum of all the minimal product-sum numbers for 2≤k≤12000?

n[k]表示minimal product-sum numbers for size=k

n[k]的上界为2*k，由于2*k总是能分解成2*k，然后2*k=k+2+(1)*(k-2)

显然n[k]的下界为k

对于一个数num 因式分解后因子个数为product 这些因子的和为sump

则须要加入的1的个数为num-sump，所以size k=num-sump+product

maxk = 12000
n=[2*maxk for i in range(maxk)]

def getpsn(num,sump,product,start):
#print(num,' ',sump,' ',product)
k = num - sump + product
if k < maxk:
if num < n[k]:
n[k] = num
for i in range(start,maxk//num * 2):   #控制num<=2*maxk
getpsn(num * i,sump + i,product + 1,i)

getpsn(1,1,1,2)
ans=sum(set(n[2:]))
print(ans)