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:
- {5, 5, 5}
- {5, 5, 1}
- {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 <bits/stdc++.h>using namespace std;int MOD = 1000000007;// Find the value of x raised to the// yth power modulo MODlong 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 theoremlong modInverse(long x){ return modPow(x, MOD - 2);}// Modular division x / y, find// modular multiplicative inverse// of y and multiply by xlong modDivision(long p, long q){ return (p * modInverse(q)) % MOD;}// Function to find Binomial Coefficient// C(n, k) in O(k) timelong 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 criteriaint 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 Codeint main(){ int N = 3, K = 5; cout << countArrays(N, K); return 0;} |
Java
// Java program for the above approachclass 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 approachMOD = 1000000007from math import sqrt# Find the value of x raised to the# yth power modulo MODdef 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 theoremdef modInverse(x): return modPow(x, MOD - 2)# Modular division x / y, find# modular multiplicative inverse# of y and multiply by xdef modDivision(p, q): return (p * modInverse(q)) % MOD# Function to find Binomial Coefficient# C(n, k) in O(k) timedef 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 criteriadef 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 Codeif __name__ == '__main__': N = 3 K = 5 print(countArrays(N, K)) # This code is contributed by SURENDRA_GANGWAR. |
C#
// C# program for the above approachusing 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
<script> // JavaScript Program to implement // the above approach let MOD = 1000000007; // Find the value of x raised to the // yth power modulo MOD function modPow(x, y) { // Stores the value of x^y let 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 function modInverse(x) { return modPow(x, MOD - 2); } // Modular division x / y, find // modular multiplicative inverse // of y and multiply by x function modDivision(p, q) { return (p * modInverse(q)) % MOD; } // Function to find Binomial Coefficient // C(n, k) in O(k) time function C(n, k) { // Base Case if (k > n) { return 0; } let p = 1, q = 1; for (let 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 function countArrays(N, K) { // Stores the resultant count of arrays let res = 1; // Find the factorization of K for (let p = 2; p <= K / p; p++) { let 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 let N = 3, K = 5; document.write(countArrays(N, K)); // This code is contributed by Potta Lokesh </script> |
Output:
3
Time Complexity: O(sqrt(N))
Auxiliary Space: O(1)
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