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描述
You are given two integers n and maxValue, which are used to describe an ideal array. A 0-indexed integer array arr of length n is considered ideal if the following conditions hold:
- Every arr[i] is a value from 1 to maxValue, for 0 <= i < n.
- Every arr[i] is divisible by arr[i - 1], for 0 < i < n.
Return the number of distinct ideal arrays of length n. Since the answer may be very large, return it modulo 10^9 + 7.
Example 1:
Input: n = 2, maxValue = 5
Output: 10
Explanation: The following are the possible ideal arrays:
- Arrays starting with the value 1 (5 arrays): [1,1], [1,2], [1,3], [1,4], [1,5]
- Arrays starting with the value 2 (2 arrays): [2,2], [2,4]
- Arrays starting with the value 3 (1 array): [3,3]
- Arrays starting with the value 4 (1 array): [4,4]
- Arrays starting with the value 5 (1 array): [5,5]
There are a total of 5 + 2 + 1 + 1 + 1 = 10 distinct ideal arrays.
Example 2:
Input: n = 5, maxValue = 3
Output: 11
Explanation: The following are the possible ideal arrays:
- Arrays starting with the value 1 (9 arrays):
- With no other distinct values (1 array): [1,1,1,1,1]
- With 2nd distinct value 2 (4 arrays): [1,1,1,1,2], [1,1,1,2,2], [1,1,2,2,2], [1,2,2,2,2]
- With 2nd distinct value 3 (4 arrays): [1,1,1,1,3], [1,1,1,3,3], [1,1,3,3,3], [1,3,3,3,3]
- Arrays starting with the value 2 (1 array): [2,2,2,2,2]
- Arrays starting with the value 3 (1 array): [3,3,3,3,3]
There are a total of 9 + 1 + 1 = 11 distinct ideal arrays.
Note:
2 <= n <= 10^4
1 <= maxValue <= 10^4
解析
根据题意，给定两个整数 n 和 maxValue ，用于描述理想数组。 如果满足以下条件，则长度为 n 的 0 索引整数数组 arr 被认为是理想的：
- 每个 arr[i] 是一个从 1 到 maxValue 的值，对于 0 <= i < n。
- 每个 arr[i] 都可以被 arr[i - 1] 整除，因为 0 < i < n。
返回长度为 n 的不同理想数组的数量。 由于答案可能非常大，因此以 10^9 + 7 为模返回。
怎么说呢，这道题有点超纲，我没有做出来，因为这道题不仅要分析清楚思路，而且考察的还有组合计算以及分解质因子等数学内容，我是看了灵神的讲解才明白，大家直接看视频吧。
解答
class Solution:
def idealArrays(self, n , maxValue ) :
mod = 10 ** 9 + 7
max_k = 13
max_n = n + max_k
dp = [[0] * (max_k + 1) for _ in range(max_n)]
dp[0][0] = 1
for i in range(1, max_n):
dp[i][0] = 1
for j in range(1, min(i, max_k) + 1):
dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j]) % mod
ks = [[] for _ in range(maxValue + 1)]
for i in range(2, maxValue + 1):
p, x = 2, i
while p * p <= x:
if x % p == 0:
k = 0
while x % p == 0:
k += 1
x //= p
ks[i].append(k)
p += 1
if x > 1:
ks[i].append(1)
res = 0
for x in range(1, maxValue + 1):
mul = 1
for k in ks[x]:
mul = mul * dp[n - 1 + k][k] % mod
res = (res + mul) % mod
return res
运行结果
47 / 47 test cases passed.
Status: Accepted
Runtime: 708 ms
Memory Usage: 20.1 MB
原题链接
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