前 N 个正整数的排列，使得素数位于素数索引 | 2套

给定一个整数N ，任务是找到前 N 个正整数的排列数，使得素数处于素数索引(对于基于 1 的索引)。
注意：由于方式的数量可能非常多，返回模 10 ⁹ + 7 的答案。
例子：

Input: N = 3 
Output: 2 
Explanation: 
Possible permutation of first 3 positive integers, such that prime numbers are at prime indices are: {1, 2, 3}, {1, 3, 2}
Input: N = 5 
Output: 12 
 

方法：使用埃拉托色尼筛

• 首先，使用埃拉托色尼筛法计算 1 到 N 之间的所有素数。
• 接下来，迭代每个位置并获得素数位置的计数，称为 k。
• 因此，对于 k 个素数，我们的选择有限，我们需要将它们排列在 k 个素数点上。
• 对于 nk 个非质数，我们的选择也很有限。我们需要将它们安排在 nk 个非主要位置。
• 这两个事件都是独立的，所以总的方式将是它们的产物。
• 在 k 个盒子中排列 k 个物体的方法数是k！

下面是上述方法的实现：

C++

// C++ program to count
// permutations from 1 to N
// such that prime numbers
// occur at prime indices
 
#include 
using namespace std;
 
static const int MOD = 1e9 + 7;
 
int numPrimeArrangements(int n)
{
    vector prime(n + 1, true);
    prime[0] = false;
    prime[1] = false;
 
    // Computing count of prime
    // numbers using sieve
    for (int i = 2; i <= sqrt(n); i++) {
        if (prime[i])
            for (int factor = 2;
                 factor * i <= n;
                 factor++)
                prime[factor * i] = false;
    }
 
    int primeIndices = 0;
    for (int i = 1; i <= n; i++)
        if (prime[i])
            primeIndices++;
 
    int mod = 1e9 + 7, res = 1;
 
    // Computing permutations for primes
    for (int i = 1; i <= primeIndices; i++)
        res = (1LL * res * i) % mod;
 
    // Computing permutations for non-primes
    for (int i = 1; i <= (n - primeIndices); i++)
        res = (1LL * res * i) % mod;
 
    return res;
}
 
// Driver program
int main()
{
    int N = 5;
    cout << numPrimeArrangements(N);
    return 0;
}

Java

// Java program to count
// permutations from 1 to N
// such that prime numbers
// occur at prime indices
  
 
import java.util.*;
 
class GFG{
  
static int MOD = (int) (1e9 + 7);
  
static int numPrimeArrangements(int n)
{
    boolean []prime = new boolean[n + 1];
    Arrays.fill(prime, true);
    prime[0] = false;
    prime[1] = false;
  
    // Computing count of prime
    // numbers using sieve
    for (int i = 2; i <= Math.sqrt(n); i++) {
        if (prime[i])
            for (int factor = 2;
                 factor * i <= n;
                 factor++)
                prime[factor * i] = false;
    }
  
    int primeIndices = 0;
    for (int i = 1; i <= n; i++)
        if (prime[i])
            primeIndices++;
  
    int mod = (int) (1e9 + 7), res = 1;
  
    // Computing permutations for primes
    for (int i = 1; i <= primeIndices; i++)
        res = (int) ((1L * res * i) % mod);
  
    // Computing permutations for non-primes
    for (int i = 1; i <= (n - primeIndices); i++)
        res = (int) ((1L * res * i) % mod);
  
    return res;
}
  
// Driver program
public static void main(String[] args)
{
    int N = 5;
    System.out.print(numPrimeArrangements(N));
}
}
 
// This code contributed by sapnasingh4991

Python3

# Python3 program to count
# permutations from 1 to N
# such that prime numbers
# occur at prime indices
import math;
 
def numPrimeArrangements(n):
     
    prime = [True for i in range(n + 1)]
     
    prime[0] = False
    prime[1] = False
     
    # Computing count of prime
    # numbers using sieve
    for i in range(2,int(math.sqrt(n)) + 1):
        if prime[i]:
            factor = 2
             
            while factor * i <= n:
                prime[factor * i] = False
                factor += 1
     
    primeIndices = 0       
    for i in range(1, n + 1):
        if prime[i]:
            primeIndices += 1
             
    mod = 1000000007
    res = 1
     
    # Computing permutations for primes
    for i in range(1, primeIndices + 1):
        res = (res * i) % mod
         
    # Computing permutations for non-primes
    for i in range(1, n - primeIndices + 1):
        res = (res * i) % mod
     
    return res
 
# Driver code       
if __name__=='__main__':
     
    N = 5
     
    print(numPrimeArrangements(N))
     
# This code is contributed by rutvik_56

C#

// C# program to count permutations
// from 1 to N such that prime numbers
// occurr at prime indices
using System;
 
class GFG{
 
//static int MOD = (int) (1e9 + 7);
 
static int numPrimeArrangements(int n)
{
    bool []prime = new bool[n + 1];
 
    for(int i = 0; i < prime.Length; i++)
       prime[i] = true;
    prime[0] = false;
    prime[1] = false;
 
    // Computing count of prime
    // numbers using sieve
    for(int i = 2; i <= Math.Sqrt(n); i++)
    {
       if (prime[i])
       {
           for(int factor = 2;
                   factor * i <= n;
                   factor++)
              prime[factor * i] = false;
       }
    }
 
    int primeIndices = 0;
    for(int i = 1; i <= n; i++)
       if (prime[i])
           primeIndices++;
 
    int mod = (int) (1e9 + 7), res = 1;
 
    // Computing permutations for primes
    for(int i = 1; i <= primeIndices; i++)
       res = (int) ((1L * res * i) % mod);
 
    // Computing permutations for non-primes
    for(int i = 1; i <= (n - primeIndices); i++)
       res = (int) ((1L * res * i) % mod);
 
    return res;
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 5;
     
    Console.Write(numPrimeArrangements(N));
}
}
 
// This code is contributed by gauravrajput1

Javascript

 
// javascript program to count
// permutations from 1 to N
// such that prime numbers
// occur at prime indices
var MOD = parseInt(1e9 + 7);
  
function numPrimeArrangements(n)
{
    var prime = Array.from({length: n+1}, (_, i) => true);
    prime[0] = false;
    prime[1] = false;
  
    // Computing count of prime
    // numbers using sieve
    for (var i = 2; i <= Math.sqrt(n); i++) {
        if (prime[i])
            for (factor = 2;
                 factor * i <= n;
                 factor++)
                prime[factor * i] = false;
    }
  
    var primeIndices = 0;
    for (var i = 1; i <= n; i++)
        if (prime[i])
            primeIndices++;
  
    var mod = parseInt( (1e9 + 7)), res = 1;
  
    // Computing permutations for primes
    for (var i = 1; i <= primeIndices; i++)
        res = ((1 * res * i) % mod);
  
    // Computing permutations for non-primes
    for (var i = 1; i <= (n - primeIndices); i++)
        res = ((1 * res * i) % mod);
  
    return res;
}
  
// Driver program
var N = 5;
document.write(numPrimeArrangements(N));
 
// This code contributed by shikhasingrajput

输出：

12

时间复杂度： O(N * log(log(N)))

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