题目
Given an array of unique integers, arr, where each integer arr[i] is strictly greater than 1.
We make a binary tree using these integers, and each number may be used for any number of times. Each non-leaf node’s value should be equal to the product of the values of its children.
Return the number of binary trees we can make. The answer may be too large so return the answer modulo 10……9 + 7.
解题思路
dp[i]表示以i为根的树的个数 dp[i] = 1 + sum(dp[n] * dp[i // n]) if i / n is integet and i / n in arr sum(dp[i] for i in arr) 为所求
代码
class Solution:
def numFactoredBinaryTrees(self, arr: List[int]) -> int:
MOD = 10**9 + 7
arr = set(arr)
@lru_cache(None)
def dp(i):
res = 1
for n in arr:
if i % n == 0 and i // n in arr:
res += dp(n) * dp(i // n)
return res
return sum(dp(i) for i in arr) % MOD
