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

Difficulty: introductory

Starter code:

class Solution:
    def numPrimeArrangements(self, n: int) -> int:
        

Solutions:
class Solution:
    def numPrimeArrangements(self, n: int) -> int:
        primes = [True] * (n + 1)
        for prime in range(2, int(math.sqrt(n)) + 1):
            if primes[prime]:
                for composite in range(prime * prime, n + 1, prime):
                    primes[composite] = False
        cnt = sum(primes[2:])
        return math.factorial(cnt) * math.factorial(n - cnt) % (10**9 + 7)        

class Solution:
    def numPrimeArrangements(self, n: int) -> int:
        # there are 25 primes from 1 to 100
        
        primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101]
        
        for i in range(len(primes)):
            if primes[i] > n:
                break
        count = i
        print(count)
        return math.factorial(count) % (10**9 + 7) * math.factorial(n - count) % (10**9 + 7) 
        


class Solution:
    def numPrimeArrangements(self, n: int) -> int:
        p_nums = []
        c_nums = []
        dp = [True for _ in range(n+1)]
        dp[0] = False
        dp[1] = False
        for i in range(2, n+1):
            if dp[i] == True:
                cnt = 2
                while i*cnt < n+1:
                    dp[i*cnt] = False
                    cnt += 1
        
        for i in range(1, n+1):
            if dp[i]:
                p_nums += [i]
            else:
                c_nums += [i]

        #return (self.fact(len(p_nums))*self.fact(len(c_nums)))%(10**9+7)  ===ok
        #模运算规则：(a * b) % m = ((a % m) * b) % m 证明方法请自行搜索
        return (self.fact(len(p_nums) % (10**9+7))*self.fact(len(c_nums)) % (10**9+7) )
    
    def fact(self, n):
        if n <= 1:
            return 1
        else:
            return n*self.fact(n-1)

        
    '''
    题目描述

中文：质数排列
英语：Prime Arrangements

解题思路

理解题意，是要求1到n的排列组合数，但并不是全排列，而是1到n中所有质数个数的阶乘和非质数个数的阶乘的乘积并对10^9 + 7即1000000007取模运算的结果。
以n=5为例，1到5中有2、3、5共3个质数，非质数则有1、4共2个，则结果为(3! * 2!) % 1000000007 = 12

代码如下

需要理解的点 模运算规则：(a * b) % m = ((a % m) * b) % m 证明方法请自行搜索

将数字分为质数和非质数两部分。
假设有 p个质数，答案就是 p!⋅(n−p)!

     '''


class Solution:
    def numPrimeArrangements(self, n: int) -> int:
        def fac(m):
            result = 1
            for i in range(1,m+1):
                result *= i
            return result
        def PrimeTell(m):
            if m == 1 or m == 4:
                return False
            if m == 2 or m == 3:
                return True
            for j in range(2, m//2):
                if m%j == 0:
                    return False
            return True
        Sum = 0
        for i in range(1,n+1):
            if PrimeTell(i):
                Sum += 1
        return fac(Sum)*fac(n-Sum)% (10**9 + 7)


class Solution:
    def numPrimeArrangements(self, n: int) -> int:
        if n == 1: return 1
        p = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]
        if n < 98:
            lo,hi = 0,24
            while lo <= hi:
                mid = (hi-lo)//2 + lo
                if p[mid] == n: 
                    idx = mid + 1
                    break
                elif p[mid-1] < n and n < p[mid]:
                    idx = mid
                    break
                elif p[mid] > n:
                    hi = mid-1
                else:
                    lo = mid+1
        else:
            idx = 25
            
        #print(idx)
        
        p_prod = 1
        for i in range(1,idx+1):
            p_prod = (p_prod*(i%1000000007))%1000000007
        
        c_prod = 1
        for i in range(1,n-idx+1):
            c_prod = (c_prod*(i%1000000007))%1000000007
                
        return (p_prod*c_prod) % 1000000007

class Solution:
    def numPrimeArrangements(self, n: int) -> int:
        primes = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,
                 61,67,71,73,79,83,89,97]
        r = sum(1 for p in primes if p <= n)
        a = 10**9 + 7
        ans = 1
        for i in range(1, min(r, n-r)+1):
            ans *= (i % a)**2
            ans %= a
        for i in range(min(r, n-r)+1, max(r, n-r)+1):
            ans *= i
            ans %= a
        return ans