Given a prime number
Note: The range of N is less than 108.
Examples:
Input: N = 13 Output: Yes Explanation: The number 13 can be written as 11 + 2, here 11 and 2 are both prime. Input: N = 11 Output: No
Simple Solution: A simple solution is to create a sieve to store all the prime numbers less than the number N. Then run a loop from 1 to N and check whether
Efficient solution: Apart from 2, all of the prime numbers are odd. So it is not possible to represent a prime number (which is odd) to be written as a sum of two odd prime numbers, so we are sure that one of the two prime number should be 2. So we have to check whether n-2 is prime or not. If it holds we print Yes else No.
For example, if the number is 19 then we have to check whether 19-2 = 17 is a prime number or not. If 17 is a prime number then print yes otherwise print no.
Below is the implementation of the above approach:
// C++ program to check if a prime number // can be expressed as sum of // two Prime Numbers #include <bits/stdc++.h> using namespace std;
// Function to check whether // a number is prime or not bool isPrime( int n)
{ if (n <= 1)
return false ;
for ( int i = 2; i <= sqrt (n); i++)
{
if (n % i == 0)
return false ;
}
return true ;
} // Function to check if a prime number // can be expressed as sum of // two Prime Numbers bool isPossible( int N)
{ // if the number is prime,
// and number-2 is also prime
if (isPrime(N) && isPrime(N - 2))
return true ;
else
return false ;
} // Driver code int main()
{ int n = 13;
if (isPossible(n))
cout << "Yes" ;
else
cout << "No" ;
return 0;
} |
Yes