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Modify N by adding its smallest positive divisor exactly K times
  • Last Updated : 01 Apr, 2021

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++




// 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




// 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

C#




// 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

Javascript




<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>
Output: 
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++14




// 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;
}
Output: 
14

 

Time Complexity: O(√N) 
Auxiliary Space: O(1)

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