# Find the number of ordered pairs such that a * p + b * q = N, where p and q are primes

Given an array arr[], and integer Q denoting number of queries and two numbers a, b, the task is to find the number of ordered pairs (p, q) such that a * p + b * q = arr[i], where p and q are prime numbers.

Examples:

Input: Q = 3, arr[] = { 2, 7, 11 }, a = 1, b = 2
Output: 0 1 2
Explanation:
2 -> There are no ordered pairs (p, q) such that p + 2*q = 2.
7 -> There is only one ordered pair (p, q) = (3, 2) such that p + 2*q = 7.
11 -> There are two ordered pairs (p, q) = (7, 2), (5, 3) such that p + 2*q = 11.

Input: Q = 2, arr[] = { 15, 25 }, a = 1, b = 2
Output: 2 3

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to store every prime number in an array using Sieve of Eratosthenes. After storing the prime numbers, count the number of ordered pairs (p, q) such that a*p + b*q = N for every combination of (p, q) in the prime array.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the number of ordered ` `// pairs such that a * p + b * q = N ` `// where p and q are primes ` ` `  `#include ` `#define size 10001 ` `using` `namespace` `std; ` `int` `prime[size]; ` `int` `freq[size]; ` ` `  `// Sieve of erastothenes ` `// to store the prime numbers ` `// and their frequency in form a*p+b*q ` `void` `sieve(``int` `a, ``int` `b) ` `{ ` `    ``prime = 1; ` ` `  `    ``// Performing Sieve of Eratosthenes ` `    ``// to find the prime numbers unto 10001 ` `    ``for` `(``int` `i = 2; i * i < size; i++) { ` `        ``if` `(prime[i] == 0) { ` `            ``for` `(``int` `j = i * 2; j < size; j += i) ` `                ``prime[j] = 1; ` `        ``} ` `    ``} ` ` `  `    ``// Loop to find the number of ` `    ``// ordered pairs for every combination ` `    ``// of the prime numbers ` `    ``for` `(``int` `p = 1; p < size; p++) { ` `        ``for` `(``int` `q = 1; q < size; q++) { ` `            ``if` `(prime[p] == 0 && prime[q] == 0 ` `                ``&& a * p + b * q < size) { ` `                ``freq[a * p + b * q]++; ` `            ``} ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `queries = 2, a = 1, b = 2; ` `    ``sieve(a, b); ` `    ``int` `arr[queries] = { 15, 25 }; ` ` `  `    ``// Printing the number of ordered pairs ` `    ``// for every query ` `    ``for` `(``int` `i = 0; i < queries; i++) { ` `        ``cout << freq[arr[i]] << ``" "``; ` `    ``} ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find the number of ordered  ` `// pairs such that a * p + b * q = N  ` `// where p and q are primes  ` `public` `class` `GFG { ` ` `  `   `  `    ``final` `static` `int` `size = ``10001``;  ` `    ``static` `int` `prime[] = ``new` `int``[size];  ` `    ``static` `int` `freq[] = ``new` `int` `[size];  ` `     `  `    ``// Sieve of erastothenes  ` `    ``// to store the prime numbers  ` `    ``// and their frequency in form a*p+b*q  ` `    ``static` `void` `sieve(``int` `a, ``int` `b)  ` `    ``{  ` `        ``prime[``1``] = ``1``;  ` `     `  `        ``// Performing Sieve of Eratosthenes  ` `        ``// to find the prime numbers unto 10001  ` `        ``for` `(``int` `i = ``2``; i * i < size; i++) {  ` `            ``if` `(prime[i] == ``0``) {  ` `                ``for` `(``int` `j = i * ``2``; j < size; j += i)  ` `                    ``prime[j] = ``1``;  ` `            ``}  ` `        ``}  ` `     `  `        ``// Loop to find the number of  ` `        ``// ordered pairs for every combination  ` `        ``// of the prime numbers  ` `        ``for` `(``int` `p = ``1``; p < size; p++) {  ` `            ``for` `(``int` `q = ``1``; q < size; q++) {  ` `                ``if` `(prime[p] == ``0` `&& prime[q] == ``0`  `                    ``&& a * p + b * q < size) {  ` `                    ``freq[a * p + b * q]++;  ` `                ``}  ` `            ``}  ` `        ``}  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `queries = ``2``, a = ``1``, b = ``2``;  ` `        ``sieve(a, b);  ` `        ``int` `arr[] = { ``15``, ``25` `};  ` `     `  `        ``// Printing the number of ordered pairs  ` `        ``// for every query  ` `        ``for` `(``int` `i = ``0``; i < queries; i++) {  ` `            ``System.out.print(freq[arr[i]] + ``" "``);  ` `        ``}  ` `     `  `     `  `    ``}  ` `} ` `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 program to find the number of ordered ` `# pairs such that a * p + b * q = N ` `# where p and q are primes ` `from` `math ``import` `sqrt ` `size ``=` `1000` `prime ``=` `[``0` `for` `i ``in` `range``(size)]  ` `freq ``=` `[``0` `for` `i ``in` `range``(size)] ` ` `  `# Sieve of erastothenes ` `# to store the prime numbers ` `# and their frequency in form a*p+b*q ` `def` `sieve(a, b): ` `    ``prime[``1``] ``=` `1` ` `  `    ``# Performing Sieve of Eratosthenes ` `    ``# to find the prime numbers unto 10001 ` `    ``for` `i ``in` `range``(``2``, ``int``(sqrt(size)) ``+` `1``, ``1``): ` `        ``if` `(prime[i] ``=``=` `0``): ` `            ``for` `j ``in` `range``(i``*``2``, size, i): ` `                ``prime[j] ``=` `1` ` `  `    ``# Loop to find the number of ` `    ``# ordered pairs for every combination ` `    ``# of the prime numbers ` `    ``for` `p ``in` `range``(``1``, size, ``1``): ` `        ``for` `q ``in` `range``(``1``, size, ``1``): ` `            ``if` `(prime[p] ``=``=` `0` `and` `prime[q] ``=``=` `0` `and` `a ``*` `p ``+` `b ``*` `q < size): ` `                ``freq[a ``*` `p ``+` `b ``*` `q] ``+``=` `1` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``queries ``=` `2` `    ``a ``=` `1` `    ``b ``=` `2` `    ``sieve(a, b) ` `    ``arr ``=` `[``15``, ``25``] ` ` `  `    ``# Printing the number of ordered pairs ` `    ``# for every query ` `    ``for` `i ``in` `range``(queries): ` `        ``print``(freq[arr[i]],end ``=` `" "``) ` ` `  `# This code is contributed by Surendra_Gangwar `

## C#

 `// C# program to find the number of ordered  ` `// pairs such that a * p + b * q = N  ` `// where p and q are primes  ` `using` `System; ` ` `  `class` `GFG {  ` ` `  `     `  `    ``static` `int` `size = 10001;  ` `    ``static` `int` `[]prime = ``new` `int``[size];  ` `    ``static` `int` `[]freq = ``new` `int` `[size];  ` `     `  `    ``// Sieve of erastothenes  ` `    ``// to store the prime numbers  ` `    ``// and their frequency in form a*p+b*q  ` `    ``static` `void` `sieve(``int` `a, ``int` `b)  ` `    ``{  ` `        ``prime = 1;  ` `     `  `        ``// Performing Sieve of Eratosthenes  ` `        ``// to find the prime numbers unto 10001  ` `        ``for` `(``int` `i = 2; i * i < size; i++) {  ` `            ``if` `(prime[i] == 0) {  ` `                ``for` `(``int` `j = i * 2; j < size; j += i)  ` `                    ``prime[j] = 1;  ` `            ``}  ` `        ``}  ` `     `  `        ``// Loop to find the number of  ` `        ``// ordered pairs for every combination  ` `        ``// of the prime numbers  ` `        ``for` `(``int` `p = 1; p < size; p++) {  ` `            ``for` `(``int` `q = 1; q < size; q++) {  ` `                ``if` `(prime[p] == 0 && prime[q] == 0 ` `                    ``&& a * p + b * q < size) {  ` `                    ``freq[a * p + b * q]++;  ` `                ``}  ` `            ``}  ` `        ``}  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main (``string``[] args)  ` `    ``{  ` `        ``int` `queries = 2, a = 1, b = 2;  ` `        ``sieve(a, b);  ` `        ``int` `[]arr = { 15, 25 };  ` `     `  `        ``// Printing the number of ordered pairs  ` `        ``// for every query  ` `        ``for` `(``int` `i = 0; i < queries; i++) {  ` `            ``Console.Write(freq[arr[i]] + ``" "``);  ` `        ``}  ` `     `  `     `  `    ``}  ` `}  ` ` `  `// This code is contributed by AnkitRai01  `

Output:

```2 3
```

Time Complexity: O(N)

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