Count of ways to represent N as sum of a prime number and twice of a square

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. N = 2*A^{2} + P, where P can be any prime number and A is any positive integer.

Note: N <= 10^{6}
Examples:

Input: N = 9
Output: 1
Explanation:
9 can be represented as sum of prime number and twice a square in only one way –
N = 9 = 7 + 2*(1^{2})

Input: N = 15
Output: 2
Explanation:
15 can be represented as sum of prime number and twice a square in two ways –
N = 15 = 7 + 2 * (2^{2})
N = 15 = 13 + 2 * (1^{2})

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++

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// 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;
}

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Output:


1





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