通过将其最小正除数正好相乘K次来修改N

给定两个正整数N和K ，任务是在每个操作中将N的值增加其最小除数超过N (超过1 )(正好是K次)后，找到N的值。

例子：

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

简单方法：最简单的方法到在范围解决这个问题，以迭代[1，K]使用变量i并且在每个操作中，找出最小的除数大于N个1和由所述最小除数大于1递增N的值的N。最后，打印N的值。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
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

// 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.

输出：

14

时间复杂度： O(K *√N)
辅助空间： O(1)

高效方法：请按照以下步骤解决问题：

• 如果N是偶数，则将N的值更新为(N + K * 2) 。
• 否则，找到大于N的1的最小正除数，例如smDiv并将值N更新为(N + smDiv +(K – 1)* 2)
• 最后，打印N的值。

下面是上述方法的实现：

C++ 14

// C++ program to implement
// the above approach
#include 
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;
}

输出：

14

时间复杂度： O(√N)
辅助空间： O(1)