Given two numbers N and K. We need to find out if ‘N’ can be written as sum of ‘K’ prime numbers.
Given N <= 10^9
Examples :
Input : N = 10 K = 2
Output : Yes
10 can be written as 5 + 5
Input : N = 2 K = 2
Output : No
The idea is to use Goldbach’s conjecture which says that every even integer (greater than 2) can be expressed as sum of two primes.
If the N = 2K and K = 1 : the answer will be Yes iff N is a prime number
If N >= 2K and K = 2 : If N is an even number answer will be Yes(Goldbach’s conjecture) and if N is odd answer will be No if N-2 is not a prime number and Yes if N-2 is a prime number. This is because we know odd + odd = even and even + odd = odd. So when N is odd, and K = 2 one number must be 2 as it is the only even prime number so now the answer depends whether N-2 is odd or not.
If N >= 2K and K >= 3 : Answer will always be Yes. When N is even N – 2*(K-2) is also even so N – 2*(K – 2) can be written as sum of two prime numbers (Goldbach’s conjecture) p, q and N can be written as 2, 2 …..K – 2 times, p, q. When N is odd N – 3 -2*(K – 3) is even so it can be written as sum of two prime numbers p, q and N can be written as 2, 2 …..K-3 times, 3, p, q
C/C++
// C++ implementation to check if N can be // written as sum of k primes #include<bits/stdc++.h> using namespace std; // Checking if a number is prime or not bool isprime(int x) { // check for numbers from 2 to sqrt(x) // if it is divisible return false for (int i=2; i*i<=x; i++) if (x%i == 0) return false; return true; } // Returns true if N can be written as sum // of K primes bool isSumOfKprimes(int N, int K) { // N < 2K directly return false if (N < 2*K) return false; // If K = 1 return value depends on primality of N if (K == 1) return isprime(N); if (K == 2) { // if N is even directly return true; if (N%2 == 0) return true; // If N is odd, then one prime must // be 2. All other primes are odd // and cannot have a pair sum as even. return isprime(N - 2); } // If K >= 3 return true; return true; } // Driver function int main() { int n = 10, k = 2; if (isSumOfKprimes (n, k)) cout << "Yes" << endl; else cout << "No" << endl; return 0; } |
Java
// Java implementation to check if N can be // written as sum of k primes public class Prime { // Checking if a number is prime or not static boolean isprime(int x) { // check for numbers from 2 to sqrt(x) // if it is divisible return false for (int i=2; i*i<=x; i++) if (x%i == 0) return false; return true; } // Returns true if N can be written as sum // of K primes static boolean isSumOfKprimes(int N, int K) { // N < 2K directly return false if (N < 2*K) return false; // If K = 1 return value depends on primality of N if (K == 1) return isprime(N); if (K == 2) { // if N is even directly return true; if (N%2 == 0) return true; // If N is odd, then one prime must // be 2. All other primes are odd // and cannot have a pair sum as even. return isprime(N - 2); } // If K >= 3 return true; return true; } public static void main (String[] args) { int n = 10, k = 2; if (isSumOfKprimes (n, k)) System.out.print("Yes"); else System.out.print("No"); } } // Contributed by Saket Kumar |
Python3
# Python implementation to check # if N can be written as sum of # k primes # Checking if a number is prime # or not def isprime(x): # check for numbers from 2 # to sqrt(x) if it is divisible # return false i = 2 while(i * i <= x): if (x % i == 0): return 0 return 1 # Returns true if N can be written # as sum of K primes def isSumOfKprimes(N, K): # N < 2K directly return false if (N < 2 * K): return 0 # If K = 1 return value depends # on primality of N if (K == 1): return isprime(N) if (K == 2): # if N is even directly # return true; if (N % 2 == 0): return 1 # If N is odd, then one # prime must be 2. All # other primes are odd # and cannot have a pair # sum as even. return isprime(N - 2); # If K >= 3 return true; return 1 # Driver function n = 10k = 2if (isSumOfKprimes (n, k)): print ("Yes") else: print ("No") # This code is Contributed by Sam007. |
C#
// C# implementation to check if N can be // written as sum of k primes using System; class GFG { // Checking if a number is prime or not static bool isprime(int x) { // check for numbers from 2 to sqrt(x) // if it is divisible return false for (int i = 2; i * i <= x; i++) if (x % i == 0) return false; return true; } // Returns true if N can be written as sum // of K primes static bool isSumOfKprimes(int N, int K) { // N < 2K directly return false if (N < 2 * K) return false; // If K = 1 return value depends on primality of N if (K == 1) return isprime(N); if (K == 2) { // if N is even directly return true; if (N % 2 == 0) return true; // If N is odd, then one prime must // be 2. All other primes are odd // and cannot have a pair sum as even. return isprime(N - 2); } // If K >= 3 return true; return true; } // Driver function public static void Main () { int n = 10, k = 2; if (isSumOfKprimes (n, k)) Console.Write("Yes"); else Console.Write("No"); } } // This code is contributed by Sam007 |
PHP
<?php // PHP implementation to check // if N can be written as sum // of k primes // Checking if a number // is prime or not function isprime($x) { // check for numbers from 2 // to sqrt(x) if it is // divisible return false for ($i = 2; $i * $i <= $x; $i++) if ($x % $i == 0) return false; return true; } // Returns true if N can be // written as sum of K primes function isSumOfKprimes($N, $K) { // N < 2K directly return false if ($N < 2 * $K) return false; // If K = 1 return value // depends on primality of N if ($K == 1) return isprime($N); if ($K == 2) { // if N is even directly // return true; if ($N % 2 == 0) return true; // If N is odd, then one prime // must be 2. All other primes // are odd and cannot have a // pair sum as even. return isprime($N - 2); } // If K >= 3 return true; return true; } // Driver Code $n = 10; $k = 2; if (isSumOfKprimes ($n, $k)) echo "Yes"; else echo"No" ; // This code is contributed by vt ?> |
Output :
Yes
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