给定一个整数N ，任务是查找前N个正整数的排列数目，以使质数处于质数索引(基于1的索引)。
注意：由于方式的数量可能非常多，请以10 9 + 7为模返回答案。
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
Explanation:
Some of the possible permutation of first 5 positive integers, such that prime numbers are at prime indices are: {1, 2, 3, 4}, {1, 3, 2, 4}, {4, 2, 3, 1}, {4, 3, 2, 1}
方法：从1到N有K个素数，从索引1到N恰有K个素数索引。因此素数的排列数为K!。同样，非素数的排列数为(NK)!。因此，排列的总数为K！*(NK)！
例如：
Given test case: [1,2,3,4,5].
2, 3 and 5 are fixed on prime index slots,
we can only swap them around.
There are 3 x 2 x 1 = 3! ways
[[2,3,5], [2,5,3], [3,2,5],
[3,5,2], [5,2,3], [5,3,2]],
For Non-prime numbers - {1,4}
[[1,4], [4,1]]
So the total is 3!*2!下面是上述方法的实现：
C++
// C++ implementation to find the
// permutation of first N positive
// integers such that prime numbers
// are at the prime indices
#define mod 1000000007
#include
using namespace std;
int fact(int n)
{
int ans = 1;
while(n > 0)
{
ans = ans * n;
n--;
}
return ans;
}
// Function to check that
// a number is prime or not
bool isPrime(int n)
{
if(n <= 1)
return false;
// Loop to check that
// number is divisible by any
// other number or not except 1
for(int i = 2; i< sqrt(n) + 1; i++)
{
if(n % i == 0)
return false;
else
return true;
}
}
// Function to find the permutations
void findPermutations(int n)
{
int prime = 0;
// Loop to find the
// number of prime numbers
for(int j = 1; j < n + 1; j++)
{
if(isPrime(j))
prime += 1;
}
// Permutation of N
// positive integers
int W = fact(prime) * fact(n - prime);
cout << (W % mod);
}
// Driver Code
int main()
{
int n = 7;
findPermutations(n);
}
// This code is contributed by Bhupendra_Singh Java
// Java implementation to find the
// permutation of first N positive
// integers such that prime numbers
// are at the prime indices
import java.io.*;
class GFG{
static int mod = 1000000007;
static int fact(int n)
{
int ans = 1;
while(n > 0)
{
ans = ans * n;
n--;
}
return ans;
}
// Function to check that
// a number is prime or not
static boolean isPrime(int n)
{
if(n <= 1)
return false;
// Loop to check that
// number is divisible by any
// other number or not except 1
for(int i = 2; i< Math.sqrt(n) + 1; i++)
{
if(n % i == 0)
return false;
else
return true;
}
return true;
}
// Function to find the permutations
static void findPermutations(int n)
{
int prime = 0;
// Loop to find the
// number of prime numbers
for(int j = 1; j < n + 1; j++)
{
if(isPrime(j))
prime += 1;
}
// Permutation of N
// positive integers
int W = fact(prime) * fact(n - prime);
System.out.println(W % mod);
}
// Driver Code
public static void main (String[] args)
{
int n = 7;
findPermutations(n);
}
}
// This code is contributed by shubhamsingh10Python3
# Python3 implementation to find the
# permutation of first N positive
# integers such that prime numbers
# are at the prime indices
import math
# Function to check that
# a number is prime or not
def isPrime(n):
if n <= 1:
return False
# Loop to check that
# number is divisible by any
# other number or not except 1
for i in range(2, int(n**0.5)+1):
if n % i == 0:
return False
else:
return True
# Constant value for modulo
CONST = int(math.pow(10, 9))+7
# Function to find the permutations
def findPermutations(n):
prime = 0
# Loop to find the
# number of prime numbers
for j in range (1, n + 1):
if isPrime(j):
prime+= 1
# Permutation of N
# positive integers
W = math.factorial(prime)*\
math.factorial(n-prime)
print (W % CONST)
# Driver Code
if __name__ == "__main__":
n = 7
# Function Call
findPermutations(n)C#
// C# implementation to find the
// permutation of first N positive
// integers such that prime numbers
// are at the prime indices
using System;
class GFG{
static int mod = 1000000007;
static int fact(int n)
{
int ans = 1;
while(n > 0)
{
ans = ans * n;
n--;
}
return ans;
}
// Function to check that
// a number is prime or not
static bool isPrime(int n)
{
if(n <= 1)
return false;
// Loop to check that
// number is divisible by any
// other number or not except 1
for(int i = 2; i < Math.Sqrt(n) + 1; i++)
{
if(n % i == 0)
return false;
else
return true;
}
return true;
}
// Function to find the permutations
static void findPermutations(int n)
{
int prime = 0;
// Loop to find the
// number of prime numbers
for(int j = 1; j < n + 1; j++)
{
if(isPrime(j))
prime += 1;
}
// Permutation of N
// positive integers
int W = fact(prime) * fact(n - prime);
Console.Write(W % mod);
}
// Driver Code
public static void Main()
{
int n = 7;
findPermutations(n);
}
}
// This code is contributed by Code_MechJavascript
输出：
144
144