Given an integer N, the task is to count the number of ways so that N can be written as the sum of a prime number and twice of a square, i.e.
, where P can be any prime number and A is any positive integer.
Note:
Examples:
Input: N = 9
Output: 1
Explanation:
9 can be represented as sum of prime number and twice a square in only one way –
Input: N = 15
Output: 2
Explanation:
15 can be represented as sum of prime number and twice a square in two ways –
[Tex]N = 15 = 13 + 2 * (1^{2}) [/Tex]
Approach: The idea is to use Seive of Eratosthenes to find all the primes and then for each prime number check for every possible number starting from 1. If any prime number and twice a square is equal to the given number then increment the count of the number of ways by 1.
Below is the implementation of the above approach:
C++
// C++ implementation to count the // number of ways a number can be // written as sum of prime number // and twice a square #include <bits/stdc++.h> using namespace std; long long int n = 500000 - 2; vector< long long int > v; // Function to mark all the // prime numbers using sieve void sieveoferanthones() { bool prime[n + 1]; // Intially all the numbers // are marked as prime memset (prime, true , sizeof (prime)); // Loop to mark the prime numbers // upto the Square root of N for ( long long int i = 2; i <= sqrt (n); i++) { if (prime[i]) for ( long long int j = i * i; j <= n; j += i) { prime[j] = false ; } } // Loop to store the prime // numbers in an array for ( long long int i = 2; i < n; i++) { if (prime[i]) v.push_back(i); } } // Function to find the number // ways to represent a number // as the sum of prime number and // square of a number void numberOfWays( long long int n) { long long int count = 0; // Loop to iterate over all the // possible prime numbers for ( long long int j = 1; 2 * ( pow (j, 2)) < n; j++) { for ( long long int i = 1; v[i] + 2 <= n; i++) { // Incrment the count if // the given number is a // valid number if (n == v[i] + (2 * ( pow (j, 2)))) count++; } } cout << count << endl; } // Driver Code int main() { sieveoferanthones(); long long int n = 9; // Function Call numberOfWays(n); return 0; } |
Java
// Java implementation to count the // number of ways a number can be // written as sum of prime number // and twice a square import java.util.*; class GFG{ static int n = 500000 - 2 ; static Vector<Integer> v = new Vector<>(); // Function to mark all the // prime numbers using sieve static void sieveoferanthones() { boolean []prime = new boolean [n + 1 ]; // Intially all the numbers // are marked as prime Arrays.fill(prime, true ); // Loop to mark the prime numbers // upto the Square root of N for ( int i = 2 ; i <= Math.sqrt(n); i++) { if (prime[i]) for ( int j = i * i; j <= n; j += i) { prime[j] = false ; } } // Loop to store the prime // numbers in an array for ( int i = 2 ; i < n; i++) { if (prime[i]) v.add(i); } } // Function to find the number // ways to represent a number // as the sum of prime number and // square of a number static void numberOfWays( int n) { int count = 0 ; // Loop to iterate over all the // possible prime numbers for ( int j = 1 ; 2 * (Math.pow(j, 2 )) < n; j++) { for ( int i = 1 ; v.get(i) + 2 <= n; i++) { // Incrment the count if // the given number is a // valid number if (n == v.get(i) + ( 2 * (Math.pow(j, 2 )))) count++; } } System.out.print(count + "\n" ); } // Driver Code public static void main(String[] args) { sieveoferanthones(); int n = 9 ; // Function Call numberOfWays(n); } } // This code is contributed by Princi Singh |
Python3
# Python3 implementation to count the # number of ways a number can be # written as sum of prime number # and twice a square import math n = 500000 - 2 v = [] # Function to mark all the # prime numbers using sieve def sieveoferanthones(): prime = [ 1 ] * (n + 1 ) # Loop to mark the prime numbers # upto the Square root of N for i in range ( 2 , int (math.sqrt(n)) + 1 ): if (prime[i] ! = 0 ): for j in range (i * i, n + 1 , i): prime[j] = False # Loop to store the prime # numbers in an array for i in range ( 2 , n): if (prime[i] ! = 0 ): v.append(i) # Function to find the number # ways to represent a number # as the sum of prime number and # square of a number def numberOfWays(n): count = 0 # Loop to iterate over all the # possible prime numbers j = 1 while ( 2 * ( pow (j, 2 )) < n): i = 1 while (v[i] + 2 < = n): # Incrment the count if # the given number is a # valid number if (n = = v[i] + ( 2 * (math. pow (j, 2 )))): count + = 1 i + = 1 j + = 1 print (count) # Driver Code sieveoferanthones() n = 9 # Function call numberOfWays(n) # This code is contributed by sanjoy_62 |
C#
// C# implementation to count the // number of ways a number can be // written as sum of prime number // and twice a square using System; using System.Collections; using System.Collections.Generic; class GFG{ static int n = 500000 - 2; static ArrayList v = new ArrayList(); // Function to mark all the // prime numbers using sieve static void sieveoferanthones() { bool []prime = new bool [n + 1]; // Intially all the numbers // are marked as prime Array.Fill(prime, true ); // Loop to mark the prime numbers // upto the Square root of N for ( int i = 2; i <= ( int )Math.Sqrt(n); i++) { if (prime[i]) { for ( int j = i * i; j <= n; j += i) { prime[j] = false ; } } } // Loop to store the prime // numbers in an array for ( int i = 2; i < n; i++) { if (prime[i]) v.Add(i); } } // Function to find the number // ways to represent a number // as the sum of prime number and // square of a number static void numberOfWays( int n) { int count = 0; // Loop to iterate over all the // possible prime numbers for ( int j = 1; 2 * (Math.Pow(j, 2)) < n; j++) { for ( int i = 1; ( int )v[i] + 2 <= n; i++) { // Incrment the count if // the given number is a // valid number if (n == ( int )v[i] + (2 * (Math.Pow(j, 2)))) count++; } } Console.Write(count); } // Driver Code public static void Main ( string [] args) { sieveoferanthones(); int n = 9; // Function call numberOfWays(n); } } // This code is contributed by rutvik_56 |
1