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Prime numbers after prime P with sum S in C++
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
#include <iostream> #include <vector> #include <cmath> using namespace std; vector<int> set; vector<int> primeNo; bool isPrimeNumber(int x) { 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++) { if (isPrimeNumber(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
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