leetcode 1175. Prime Arrangements 解法 python

一.问题描述

Return the number of permutations of 1 to n so that prime numbers are at prime indices (1-indexed.)

(Recall that an integer is prime if and only if it is greater than 1, and cannot be written as a product of two positive integers both smaller than it.)

Since the answer may be large, return the answer modulo 10^9 + 7.

Example 1:

Input: n = 5
Output: 12
Explanation: For example [1,2,5,4,3] is a valid permutation, but [5,2,3,4,1] is not because the prime number 5 is at index 1.

Example 2:

Input: n = 100
Output: 682289015

Constraints:

1 <= n <= 100

二.解决思路

其实就是找1～n有多少个质数，然后计算所有质数的全排列和所有非质数的全排列然后相乘求余

本身这个东西是比较简单的，

你可以一个个迭代判断每个数是不是质数，

但是求一定范围内质数个数的最快算法是（除了查表法）Sieve of Eratosthenes（埃拉托色泥筛选法）算法

其实根据定理：一个合数总是可以分解成若干个质数的乘积，那么如果把质数（最初只知道2是质数）的倍数都去掉，那么剩下的就是质数了。

所以我们可以从2开始一个个的把倍数都去掉，直到大于n为止。

给出两种实现

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三.源码

1.一个个的求是否是质数

from functools import reduce
import math
class Solution:def numPrimeArrangements(self, n: int) -> int:cnt_prime=0for i in range(2,n+1):is_prime=1for t in range(2,int(math.sqrt(i))+1):if i%t==0:is_prime=0breakcnt_prime+=is_primeif cnt_prime!=0: # reduce函数用来求阶乘per_prime=reduce(lambda x,y:x*y, range(1,cnt_prime+1))else:per_prime=1return (per_prime*(reduce(lambda x,y:x*y, range(1,n-cnt_prime+1))))%(pow(10,9)+7)

2.埃拉托色泥筛选法

from functools import reduce
class Solution:def numPrimeArrangements(self, n: int) -> int:is_prime=[1]*(n+1)is_prime[0],is_prime[1],=0,0p=2while p*p<=n:if is_prime[p]:for t in range(p*p,n+1,p):is_prime[t]=0p+=1cnt_prime=sum(is_prime)if cnt_prime!=0:per_prime=reduce(lambda x,y:x*y, range(1,cnt_prime+1))else:per_prime=1return (per_prime*(reduce(lambda x,y:x*y, range(1,n-cnt_prime+1))))%(10**9+7)