Given two positive integers N and K, the task is to find the value of N after incrementing the value of N in each operation by its smallest divisor exceeding N ( exceeding 1 ), exactly K times.
Examples:
Input: N = 5, K = 2
Output: 12
Explanation:
Smallest divisor of N (= 5) is 5. Therefore, N = 5 + 5 = 10.
Smallest divisor of N (= 10) is 2. Therefore, N = 5 + 2 = 12.
Therefore, the required output is 12.Input: N = 6, K = 4
Output: 14
Naive Approach: The simplest approach to solve this problem to iterate over the range [1, K] using variable i and in each operation, find the smallest divisors greater than 1 of N and increment the value of N by the smallest divisor greater than 1 of N. Finally, print the value of N.
Below is the implementation of the above approach:
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the smallest // divisor of N greater than 1 int smallestDivisorGr1( int N)
{ for ( int i = 2; i <= sqrt (N);
i++) {
// If i is a divisor
// of N
if (N % i == 0) {
return i;
}
}
// If N is a prime number
return N;
} // Function to find the value of N by // performing the operations K times int findValOfNWithOperat( int N, int K)
{ // Iterate over the range [1, K]
for ( int i = 1; i <= K; i++) {
// Update N
N += smallestDivisorGr1(N);
}
return N;
} // Driver Code int main()
{ int N = 6, K = 4;
cout << findValOfNWithOperat(N, K);
return 0;
} |
// Java program to implement // the above approach class GFG{
// Function to find the smallest // divisor of N greater than 1 static int smallestDivisorGr1( int N)
{ for ( int i = 2 ; i <= Math.sqrt(N);
i++) {
// If i is a divisor
// of N
if (N % i == 0 ) {
return i;
}
}
// If N is a prime number
return N;
} // Function to find the value of N by // performing the operations K times static int findValOfNWithOperat( int N, int K)
{ // Iterate over the range [1, K]
for ( int i = 1 ; i <= K; i++)
{
// Update N
N += smallestDivisorGr1(N);
}
return N;
} // Driver Code public static void main(String[] args)
{ int N = 6 , K = 4 ;
System.out.print(findValOfNWithOperat(N, K));
} } // This code is contributed by shikhasingrajput |
# Python 3 program to implement # the above approach import math
# Function to find the smallest # divisor of N greater than 1 def smallestDivisorGr1(N):
for i in range ( 2 , int (math.sqrt(N)) + 1 ):
# If i is a divisor
# of N
if (N % i = = 0 ):
return i
# If N is a prime number
return N
# Function to find the value of N by # performing the operations K times def findValOfNWithOperat(N, K):
# Iterate over the range [1, K]
for i in range ( 1 , K + 1 ):
# Update N
N + = smallestDivisorGr1(N)
return N
# Driver Code if __name__ = = "__main__" :
N = 6
K = 4
print (findValOfNWithOperat(N, K))
# This code is contributed by ukasp.
|
// C# program to implement // the above approach using System;
public class GFG
{ // Function to find the smallest
// divisor of N greater than 1
static int smallestDivisorGr1( int N)
{
for ( int i = 2; i <= Math.Sqrt(N);
i++) {
// If i is a divisor
// of N
if (N % i == 0) {
return i;
}
}
// If N is a prime number
return N;
}
// Function to find the value of N by
// performing the operations K times
static int findValOfNWithOperat( int N, int K)
{
// Iterate over the range [1, K]
for ( int i = 1; i <= K; i++)
{
// Update N
N += smallestDivisorGr1(N);
}
return N;
}
// Driver Code
public static void Main(String[] args)
{
int N = 6, K = 4;
Console.Write(findValOfNWithOperat(N, K));
}
} // This code is contributed by 29AjayKumar |
<script> // JavaScript program to implement // the above approach // Function to find the smallest // divisor of N greater than 1 function smallestDivisorGr1(N)
{ for (let i = 2; i <= Math.sqrt(N);
i++) {
// If i is a divisor
// of N
if (N % i == 0)
{
return i;
}
}
// If N is a prime number
return N;
} // Function to find the value of N by // performing the operations K times function findValOfNWithOperat(N, K)
{ // Iterate over the range [1, K]
for (let i = 1; i <= K; i++)
{
// Update N
N += smallestDivisorGr1(N);
}
return N;
} // Driver Code let N = 6, K = 4; document.write(findValOfNWithOperat(N, K)); // This code is contributed by Surbhi Tyagi. </script> |
14
Time Complexity: O(K * ?N)
Auxiliary Space: O(1)
Efficient Approach: Follow the steps below to solve the problem:
- If N is an even number then update the value of N to (N + K * 2).
- Otherwise, find the smallest positive divisor greater than 1 of N say, smDiv and update the value N to (N + smDiv + (K – 1) * 2)
- Finally, print the value of N.
Below is the implementation of the above approach:
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the smallest // divisor of N greater than 1 int smallestDivisorGr1( int N)
{ for ( int i = 2; i <= sqrt (N);
i++) {
// If i is a divisor
// of N
if (N % i == 0) {
return i;
}
}
// If N is a prime number
return N;
} // Function to find the value of N by // performing the operations K times int findValOfNWithOperat( int N, int K)
{ // If N is an even number
if (N % 2 == 0) {
// Update N
N += K * 2;
}
// If N is an odd number
else {
// Update N
N += smallestDivisorGr1(N)
+ (K - 1) * 2;
}
return N;
} // Driver Code int main()
{ int N = 6, K = 4;
cout << findValOfNWithOperat(N, K);
return 0;
} |
// Java program to implement // the above approach class GFG{
// Function to find the smallest // divisor of N greater than 1 static int smallestDivisorGr1( int N)
{ for ( int i = 2 ; i <= Math.sqrt(N); i++)
{
// If i is a divisor
// of N
if (N % i == 0 )
{
return i;
}
}
// If N is a prime number
return N;
} // Function to find the value of N by // performing the operations K times static int findValOfNWithOperat( int N, int K)
{ // If N is an even number
if (N % 2 == 0 )
{
// Update N
N += K * 2 ;
}
// If N is an odd number
else
{
// Update N
N += smallestDivisorGr1(N) + (K - 1 ) * 2 ;
}
return N;
} // Driver Code public static void main(String[] args)
{ int N = 6 , K = 4 ;
System.out.print(findValOfNWithOperat(N, K));
} } // This code is contributed by target_2 |
# Python program to implement # the above approach # Function to find the smallest # divisor of N greater than 1 def smallestDivisorGr1(N):
for i in range (sqrt(N)):
i + = 1
# If i is a divisor
# of N
if (N % i = = 0 ):
return i
# If N is a prime number
return N
# Function to find the value of N by # performing the operations K times def findValOfNWithOperat(N, K):
# If N is an even number
if (N % 2 = = 0 ):
# Update N
N + = K * 2
# If N is an odd number
else :
# Update N
N + = smallestDivisorGr1(N) + (K - 1 ) * 2
return N
# Driver Code N = 6
K = 4
print (findValOfNWithOperat(N, K))
# This code is contributed by shivanisinghss2110 |
// C# program to implement // the above approach using System;
class GFG{
// Function to find the smallest // divisor of N greater than 1 static int smallestDivisorGr1( int N)
{ for ( int i = 2; i <= Math.Sqrt(N); i++)
{
// If i is a divisor
// of N
if (N % i == 0)
{
return i;
}
}
// If N is a prime number
return N;
} // Function to find the value of N by // performing the operations K times static int findValOfNWithOperat( int N, int K)
{ // If N is an even number
if (N % 2 == 0)
{
// Update N
N += K * 2;
}
// If N is an odd number
else
{
// Update N
N += smallestDivisorGr1(N) + (K - 1) * 2;
}
return N;
} // Driver code static public void Main()
{ int N = 6, K = 4;
Console.Write(findValOfNWithOperat(N, K));
} } // This code is contributed by Khushboogoyal499 |
<script> // Function to find the smallest // divisor of N greater than 1 function smallestDivisorGr1( N)
{ for ( var i = 2; i <= Math.sqrt(N);
i++) {
// If i is a divisor
// of N
if (N % i == 0) {
return i;
}
}
// If N is a prime number
return N;
} // Function to find the value of N by // performing the operations K times function findValOfNWithOperat(N, K)
{ // If N is an even number
if (N % 2 == 0) {
// Update N
N += K * 2;
}
// If N is an odd number
else {
// Update N
N += smallestDivisorGr1(N)
+ (K - 1) * 2;
}
return N;
} // Driver Code var N = 6, K = 4;
document.write(findValOfNWithOperat(N, K));
</script> |
14
Time Complexity: O(?N)
Auxiliary Space: O(1)