Count of N-size Arrays that can be formed starting with K such that every element is divisible by next

• Last Updated : 25 Jan, 2022

Given two integers N and K, the task is to find the number of different arrays of size N that can be formed having the first element as K such that every element except last, is divisible by the next element in the array. Since the count can be very large, so print the modulo 109 + 7.

Examples:

Input: N = 3, K = 5
Output: 3
Explanation:
There are 3 possible valid array starting with value 5 satisfying the given criteria:

1. {5, 5, 5}
2. {5, 5, 1}
3. {5, 1, 1}.

Therefore the total count is 3.

Input: N = 3, K = 6
Output: 9

Approach: The given problem can be solved by using the Number Theory and Combinatorics. The first element of the array is K, then the next array element is one of the factors of the K. Follow the steps below to solve the given problem:

• Initialize a variable, say res as 1 that stores the resultant count of arrays formed.
• Find all the powers of prime factors of the number K and for each prime factor P perform the following steps:
• Find the number of times P occurs in the value K. Let that count be count.
• The number of possible ways to keep one of the factors of P is given by (N – count + 1)Ccount.
• Multiply the value (N – count + 1)Ccount with the value res.
• After completing the above steps, print the value of res as the result.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach   #include using namespace std; int MOD = 1000000007;   // Find the value of x raised to the // yth power modulo MOD long modPow(long x, long y) {     // Stores the value of x^y     long r = 1, a = x;       // Iterate until y is positive     while (y > 0) {         if ((y & 1) == 1) {             r = (r * a) % MOD;         }         a = (a * a) % MOD;         y /= 2;     }       return r; } // Function to perform the Modular // Multiplicative Inverse using the // Fermat's little theorem long modInverse(long x) {     return modPow(x, MOD - 2); }   // Modular division x / y, find // modular multiplicative inverse // of y and multiply by x long modDivision(long p, long q) {     return (p * modInverse(q)) % MOD; }   // Function to find Binomial Coefficient // C(n, k) in O(k) time long C(long n, int k) {     // Base Case     if (k > n) {         return 0;     }     long p = 1, q = 1;       for (int i = 1; i <= k; i++) {           // Update the value of p and q         q = (q * i) % MOD;         p = (p * (n - i + 1)) % MOD;     }       return modDivision(p, q); }   // Function to find the count of arrays // having K as the first element satisfying // the given criteria int countArrays(int N, int K) {     // Stores the resultant count of arrays     long res = 1;       // Find the factorization of K     for (int p = 2; p <= K / p; p++) {           int c = 0;           // Stores the count of the exponent         // of the currentprime factor         while (K % p == 0) {             K /= p;             c++;         }         res = (res * C(N - 1 + c, c))               % MOD;     }     if (N > 1) {           // N is one last prime factor,         // for c = 1 -> C(N-1+1, 1) = N         res = (res * N) % MOD;     }       // Return the total count     return res; }   // Driver Code int main() {     int N = 3, K = 5;     cout << countArrays(N, K);       return 0; }

Java

 // Java program for the above approach class GFG {       public static int MOD = 1000000007;       // Find the value of x raised to the     // yth power modulo MOD     public static long modPow(long x, long y)     {                 // Stores the value of x^y         long r = 1, a = x;           // Iterate until y is positive         while (y > 0) {             if ((y & 1) == 1) {                 r = (r * a) % MOD;             }             a = (a * a) % MOD;             y /= 2;         }           return r;     }       // Function to perform the Modular     // Multiplicative Inverse using the     // Fermat's little theorem     public static long modInverse(long x) {         return modPow(x, MOD - 2);     }       // Modular division x / y, find     // modular multiplicative inverse     // of y and multiply by x     public static long modDivision(long p, long q) {         return (p * modInverse(q)) % MOD;     }       // Function to find Binomial Coefficient     // C(n, k) in O(k) time     public static long C(long n, int k) {         // Base Case         if (k > n) {             return 0;         }         long p = 1, q = 1;           for (int i = 1; i <= k; i++) {               // Update the value of p and q             q = (q * i) % MOD;             p = (p * (n - i + 1)) % MOD;         }           return modDivision(p, q);     }       // Function to find the count of arrays     // having K as the first element satisfying     // the given criteria     public static long countArrays(int N, int K) {         // Stores the resultant count of arrays         long res = 1;           // Find the factorization of K         for (int p = 2; p <= K / p; p++) {               int c = 0;               // Stores the count of the exponent             // of the currentprime factor             while (K % p == 0) {                 K /= p;                 c++;             }             res = (res * C(N - 1 + c, c)) % MOD;         }         if (N > 1) {               // N is one last prime factor,             // for c = 1 -> C(N-1+1, 1) = N             res = (res * N) % MOD;         }           // Return the total count         return res;     }       // Driver Code     public static void main(String args[]) {         int N = 3, K = 5;         System.out.println(countArrays(N, K));       } }   // This code is contributed by saurabh_jaiswal.

Python3

 # python 3 program for the above approach MOD = 1000000007   from math import sqrt   # Find the value of x raised to the # yth power modulo MOD def modPow(x, y):         # Stores the value of x^y     r = 1     a = x       # Iterate until y is positive     while(y > 0):         if ((y & 1) == 1):             r = (r * a) % MOD         a = (a * a) % MOD         y /= 2       return r # Function to perform the Modular # Multiplicative Inverse using the # Fermat's little theorem def modInverse(x):     return modPow(x, MOD - 2)   # Modular division x / y, find # modular multiplicative inverse # of y and multiply by x def modDivision(p, q):     return (p * modInverse(q)) % MOD   # Function to find Binomial Coefficient # C(n, k) in O(k) time def C(n, k):     # Base Case     if (k > n):         return 0       p = 1     q = 1       for i in range(1,k+1,1):           # Update the value of p and q         q = (q * i) % MOD         p = (p * (n - i + 1)) % MOD       return modDivision(p, q)   # Function to find the count of arrays # having K as the first element satisfying # the given criteria def countArrays(N, K):     # Stores the resultant count of arrays     res = 1       # Find the factorization of K     for p in range(2,int(sqrt(K)),1):         c = 0           # Stores the count of the exponent         # of the currentprime factor         while (K % p == 0):             K /= p             c += 1         res = (res * C(N - 1 + c, c)) % MOD     if (N > 1):           # N is one last prime factor,         # for c = 1 -> C(N-1+1, 1) = N         res = (res * N) % MOD       # Return the total count     return res   # Driver Code if __name__ == '__main__':     N = 3     K = 5     print(countArrays(N, K))           # This code is contributed by SURENDRA_GANGWAR.

C#

 // C# program for the above approach using System; public class GFG {          public static int MOD = 1000000007;       // Find the value of x raised to the     // yth power modulo MOD     public static long modPow(long x, long y)     {                 // Stores the value of x^y         long r = 1, a = x;           // Iterate until y is positive         while (y > 0) {             if ((y & 1) == 1) {                 r = (r * a) % MOD;             }             a = (a * a) % MOD;             y /= 2;         }           return r;     }       // Function to perform the Modular     // Multiplicative Inverse using the     // Fermat's little theorem     public static long modInverse(long x) {         return modPow(x, MOD - 2);     }       // Modular division x / y, find     // modular multiplicative inverse     // of y and multiply by x     public static long modDivision(long p, long q) {         return (p * modInverse(q)) % MOD;     }       // Function to find Binomial Coefficient     // C(n, k) in O(k) time     public static long C(long n, int k) {         // Base Case         if (k > n) {             return 0;         }         long p = 1, q = 1;           for (int i = 1; i <= k; i++) {               // Update the value of p and q             q = (q * i) % MOD;             p = (p * (n - i + 1)) % MOD;         }           return modDivision(p, q);     }       // Function to find the count of arrays     // having K as the first element satisfying     // the given criteria     public static long countArrays(int N, int K) {         // Stores the resultant count of arrays         long res = 1;           // Find the factorization of K         for (int p = 2; p <= K / p; p++) {               int c = 0;               // Stores the count of the exponent             // of the currentprime factor             while (K % p == 0) {                 K /= p;                 c++;             }             res = (res * C(N - 1 + c, c)) % MOD;         }         if (N > 1) {               // N is one last prime factor,             // for c = 1 -> C(N-1+1, 1) = N             res = (res * N) % MOD;         }           // Return the total count         return res;     }       // Driver Code     static public void Main (){           int N = 3, K = 5;         Console.WriteLine(countArrays(N, K));     } }   // This code is contributed by Dharanendra L V.

Javascript



Output:

3

Time Complexity: O(sqrt(N))
Auxiliary Space: O(1)

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