K重复添加N的最小除数后形成的数字

给定两个整数N和K ，任务是生成执行K次运算的最终结果，其中涉及在每一步中将N的当前值除1以外的最小除数。
例子：

Input: N = 9, K = 4 
Output: 18 
Explanation: 
Divisors of 9 are {1, 3, 9} 
1st Operation: N = 9 + 3 = 12 
2nd Operation: N = 12 + 2 = 14 
3rd Operation: N = 14 + 2 = 16 
4th Operation: N = 16 + 2 = 18
Input: N = 16, K = 3 
Output: 22 
 

天真的方法：蛮力解决此问题的方法是执行K次操作，然后打印最终数字。
高效方法：这里的技巧是，如果给定的N为偶数，则所有K个运算的最小除数始终为2。因此，所需的^第K个数将很简单

required number = N + K * 2

同样，如果N为奇数，则最小除数将为奇数。因此，将它们相加将得出一个偶数值(奇数+奇数=偶数)。因此，现在可以将上述技巧应用于(K-1)操作，即

required number = N + smallest divisor of N + (K – 1) * 2

下面是上述方法的实现：

C++

// C++ program to find the Kth number
// formed after repeated addition of
// smallest divisor of N
#include
#include 
using namespace std;
 
void FindValue(int N, int K)
{
     
    // If N is even
    if( N % 2 == 0 )
    {
        N = N + 2 * K;
    }
     
    // If N is odd
    else
    {
        int i;
         
        for(i = 2; i < sqrt(N) + 1; i++)
        {
           if(N % i == 0)
              break;
        }
             
        // Add smallest divisor to N
        N = N + i;
         
        // Updated N is even
        N = N + 2 * ( K - 1 );
    }
    cout << N << endl;
}
     
// Driver code
int main()
{
    int N = 9;
    int K = 4;
     
    FindValue( N, K );
}
     
// This code is contributed by Surendra_Gangwar

Java

// Java program to find the Kth number
// formed after repeated addition of
// smallest divisor of N
import java.util.*;
class GFG{
 
static void FindValue(int N, int K)
{
     
    // If N is even
    if( N % 2 == 0 )
    {
        N = N + 2 * K;
    }
     
    // If N is odd
    else
    {
        int i;
         
        for(i = 2; i < Math.sqrt(N) + 1; i++)
        {
            if(N % i == 0)
                break;
        }
             
        // Add smallest divisor to N
        N = N + i;
         
        // Updated N is even
        N = N + 2 * ( K - 1 );
    }
    System.out.print(N);
}
     
// Driver code
public static void main(String args[])
{
    int N = 9;
    int K = 4;
     
    FindValue( N, K );
}
}
 
// This code is contributed by Nidhi_biet

Python3

# Python3 program to find the
# Kth number formed after
# repeated addition of
# smallest divisor of N
 
import math
 
def FindValue(N, K):
     
    # If N is even
    if( N % 2 == 0 ):
        N = N + 2 * K
     
    # If N is odd
    else:
         
        # Find the smallest divisor
        for i in range( 2, (int)(math.sqrt(N))+1 ):
            if( N % i == 0):
                break
             
        # Add smallest divisor to N
        N = N + i
         
        # Updated N is even
        N = N + 2 * ( K - 1 )
 
    print(N)
 
# Driver code
if __name__ == "__main__":
    N = 9
    K = 4
    FindValue( N, K )

C#

// C# program to find the Kth number
// formed after repeated addition of
// smallest divisor of N
using System;
class GFG{
 
static void FindValue(int N, int K)
{
     
    // If N is even
    if( N % 2 == 0 )
    {
        N = N + 2 * K;
    }
     
    // If N is odd
    else
    {
        int i;
         
        for(i = 2; i < Math.Sqrt(N) + 1; i++)
        {
            if(N % i == 0)
                break;
        }
             
        // Add smallest divisor to N
        N = N + i;
         
        // Updated N is even
        N = N + 2 * ( K - 1 );
    }
    Console.WriteLine(N);
}
     
// Driver code
public static void Main()
{
    int N = 9;
    int K = 4;
     
    FindValue( N, K );
}
}
 
// This code is contributed by Code_Mech

Javascript

// JavaScript program to find the Kth number
// formed after repeated addition of
// smallest divisor of N
function FindValue(N, K)
{
     
    // If N is even
    if( N % 2 == 0 )
    {
        N = N + 2 * K;
    }
     
    // If N is odd
    else
    {
        let i;
         
        for(i = 2; i < Math.sqrt(N) + 1; i++)
        {
        if(N % i == 0)
            break;
        }
             
        // Add smallest divisor to N
        N = N + i;
         
        // Updated N is even
        N = N + 2 * ( K - 1 );
    }
    document.write(N + "<br>");
}
     
// Driver code
let N = 9;
let K = 4;
     
FindValue( N, K );
 
// This code is contributed by Surbhi Tyagi.

输出：

18

时间复杂度： O(√N)