Given an integer N, the task is to represent N as the sum of two composite integers. There can be multiple ways possible, print any one of them. If it is not possible to represent the number as the sum of two composite numbers then print -1.
Examples:
Input: N = 13
Output: 4 9
4 + 9 = 13 and both 4 and 9 are composite.
Input: N = 18
Output: 4 14
Approach: When N ? 11 then only 8 and 10 are the integers which can be represented as the sum of two composite integers i.e. 4 + 4 and 4 + 6 respectively.
When N > 11 then there are two cases:
- When N is even: N can be represented as 4 + (N – 4) since both are composite.
- When N is odd: N can be represented as 9 + (N – 9).
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to find two composite // numbers which when added // give sum as n void findNums( int n)
{ // Only 8 and 10 can be represented
// as the sum of two composite integers
if (n <= 11) {
if (n == 8)
cout << "4 4" ;
if (n == 10)
cout << "4 6" ;
else
cout << "-1" ;
return ;
}
// If n is even
if (n % 2 == 0)
cout << "4 " << (n - 4);
// If n is odd
else
cout << "9 " << (n - 9);
} // Driver code int main()
{ int n = 13;
findNums(n);
return 0;
} |
// Java implementation of the approach class GFG
{ // Function to find two composite // numbers which when added // give sum as n static void findNums( int n)
{ // Only 8 and 10 can be represented
// as the sum of two composite integers
if (n <= 11 )
{
if (n == 8 )
System.out.print( "4 4" );
if (n == 10 )
System.out.print( "4 6" );
else
System.out.print( "-1" );
return ;
}
// If n is even
if (n % 2 == 0 )
System.out.print( "4 " + (n - 4 ));
// If n is odd
else
System.out.print( "9 " + (n - 9 ));
} // Driver code public static void main(String args[])
{ int n = 13 ;
findNums(n);
} } // This code is contributed by andrew1234 |
# Python3 implementation of the approach # Function to find two composite # numbers which when added # give sum as n def findNums(n):
# Only 8 and 10 can be represented
# as the sum of two composite integers
if (n < = 11 ):
if (n = = 8 ):
print ( "4 4" , end = " " )
if (n = = 10 ):
print ( "4 6" , end = " " )
else :
print ( "-1" , end = " " )
# If n is even
if (n % 2 = = 0 ):
print ( "4 " , (n - 4 ), end = " " )
# If n is odd
else :
print ( "9 " , n - 9 , end = " " )
# Driver code n = 13
findNums(n) # This code is contributed by Mohit Kumar |
// C# implementation of the approach using System;
class GFG
{ // Function to find two composite
// numbers which when added
// give sum as n
static void findNums( int n)
{
// Only 8 and 10 can be represented
// as the sum of two composite integers
if (n <= 11)
{
if (n == 8)
Console.Write( "4 4" );
if (n == 10)
Console.Write( "4 6" );
else
Console.Write( "-1" );
return ;
}
// If n is even
if (n % 2 == 0)
Console.Write( "4 " + (n - 4));
// If n is odd
else
Console.Write( "9 " + (n - 9));
}
// Driver code
public static void Main()
{
int n = 13;
findNums(n);
}
} // This code is contributed by AnkitRai01 |
<script> // javascript implementation of the approach // Function to find two composite // numbers which when added // give sum as n function findNums(n)
{ // Only 8 and 10 can be represented
// as the sum of two composite integers
if (n <= 11)
{
if (n == 8)
document.write( "4 4" );
if (n == 10)
document.write( "4 6" );
else
document.write( "-1" );
return ;
}
// If n is even
if (n % 2 == 0)
document.write( "4 " + (n - 4));
// If n is odd
else
document.write( "9 " + (n - 9));
} // Driver code var n = 13;
findNums(n); // This code contributed by shikhasingrajput </script> |
9 4
Time Complexity: O(1)
Auxiliary Space: O(1)