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]
[Tex]N = 15 = 13 + 2 * (1^{2}) [/Tex]
Approach: The idea is to use Sieve 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 sievevoid sieveoferanthones(){ bool prime[n + 1]; // Initially 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 numbervoid 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++) { // Increment the count if // the given number is a // valid number if (n == v[i] + (2 * (pow(j, 2)))) count++; } } cout << count << endl;}// Driver Codeint 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 squareimport 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]; // Initially 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++) { // Increment 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 squareimport mathn = 500000 - 2v = []# Function to mark all the# prime numbers using sievedef 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 numberdef 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): # Increment 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 Codesieveoferanthones()n = 9# Function callnumberOfWays(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 squareusing 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]; // Initially 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++) { // Increment 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 |
Javascript
<script>// JavaScript implementation to count the// number of ways a number can be// written as sum of prime number// and twice a squarelet n = 500000 - 2;let v = []; // Function to mark all the// prime numbers using sievefunction sieveoferanthones(){ let prime = Array.from({length: n+1}, (_, i) => true); // Loop to mark the prime numbers // upto the Square root of N for (let i = 2; i <= Math.sqrt(n); i++) { if (prime[i]) for (let j = i * i; j <= n; j += i) { prime[j] = false; } } // Loop to store the prime // numbers in an array for (let i = 2; i < n; i++) { if (prime[i]) v.push(i); }} // Function to find the number// ways to represent a number// as the sum of prime number and// square of a numberfunction numberOfWays(n){ let count = 0; // Loop to iterate over all the // possible prime numbers for (let j = 1; 2 * (Math.pow(j, 2)) < n; j++) { for (let i = 1; v[i] + 2 <= n; i++) { // Increment the count if // the given number is a // valid number if (n == v[i] + (2 * (Math.pow(j, 2)))) count++; } } document.write(count + "<br/>");} // Driver Code sieveoferanthones(); let N = 9; // Function Call numberOfWays(N); </script> |
Output:
1
Time Complexity: O(n2)
Auxiliary Space: O(n)
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