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 1int 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 timesint 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 Codeint main(){ int N = 6, K = 4; cout << findValOfNWithOperat(N, K); return 0;} |
Java
// Java program to implement// the above approachclass GFG{// Function to find the smallest// divisor of N greater than 1static 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 timesstatic 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 Codepublic 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 approachusing 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 |
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 1int 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 timesint 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 Codeint main(){ int N = 6, K = 4; cout << findValOfNWithOperat(N, K); return 0;} |
14
Time Complexity: O(√N)
Auxiliary Space: O(1)
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