# Prime numbers after prime P with sum S in C++

C++Server Side ProgrammingProgramming

In this problem, we are given three numbers, sum S, prime P, and N. Our task is to find all N prime numbers greater than P whose sum is equal to S.

Let’s take an example to understand our problem

Input: N = 2, P = 5, S = 18
Output: 7 11
Explanation: Prime numbers greater than 5 : 7 11 13
Sum = 7 + 11 = 18

To solve this problem, we have to find all prime numbers between P and S. And then find N prime numbers which sum up to S. For this we will use backtracking.

Program to show the implementation of our solution

## Example

Live Demo

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
vector<int> set;
vector<int> primeNo;
int sqroot = sqrt(x);
bool flag = true;
if (x == 1)
return false;
for (int i = 2; i <= sqroot; i++)
if (x % i == 0)
return false;
return true;
}
void printPrimes() {
int length = set.size();
for (int i=0; i<length; i++)
cout<<set[i]<<"\t";
cout<<endl;
}
void GeneratePrimeSum(int total, int N, int S, int index) {
if (total == S && set.size() == N) {
printPrimes();
return;
}
if (total > S || index == primeNo.size())
return;
set.push_back(primeNo[index]);
GeneratePrimeSum(total+primeNo[index], N, S, index+1);
set.pop_back();
GeneratePrimeSum(total, N, S, index+1);
}
void PrimesWithSum(int N, int S, int P) {
for (int i = P+1; i <=S ; i++) {
primeNo.push_back(i);
}
if (primeNo.size() < N)
return;
GeneratePrimeSum(0, N, S, 0);
}
int main() {
int S = 23, N = 3, P = 3;
cout<<N<<" Prime numbers greater than "<<P<<" with sum = "<<S<<" are :\n";
PrimesWithSum(N, S, P);
return 0;
}

## Output

3 Prime numbers greater than 3 with sum = 23 are :
5   7   11