将X转换为Y的最大素数移动

给定两个整数X和Y ，任务是使用以下操作将X转换为Y ：

1. 将任何质数加到X。
2. 从Y减去任何质数。

打印所需的最大此类操作数；如果无法将X转换为Y ，则打印-1 。

例子：

Input: X = 2, Y = 4
Output: 1
2 -> 4

Input: X = 5, Y = 6
Output: -1
It is impossible to convert 5 to 6
with the given operations.

方法：由于任务是使操作最大化，因此必须在每个操作中将最小可能值添加到X。由于该值必须是质数，因此可以使用最少的两个质数，即2和3，因为它们都是质数，并且可以覆盖偶数和奇数奇偶校验。现在，有三种情况：

• 如果X> Y，则答案将为-1，因为在给定的操作下X不能等于Y。
• 如果X = Y，则答案将为0 。
• 如果X 则计算P = Y – X，并且，
  • 如果P = 1，则答案将为-1，因为1不是质数，因此无法相加或相减。
  • 如果P为偶数，则可以将2重复添加到X，答案将为P / 2
  • 如果P等于偶数，则在X上加3 ，然后可以再次将2重复添加到新的X上，使其等于Y ，这种情况下的结果将是1 +((P – 3)/ 2) 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the maximum operations
// required to convert X to Y
int maxOperations(int X, int Y)
{
  
    // X cannot be converted to Y
    if (X > Y)
        return -1;
  
    int diff = Y - X;
  
    // If the differecne is 1
    if (diff == 1)
        return -1;
  
    // If the difference is even
    if (diff % 2 == 0)
        return (diff / 2);
  
    // Add 3 to X and the new
    // difference will be even
    return (1 + ((diff - 3) / 2));
}
  
// Driver code
int main()
{
    int X = 5, Y = 16;
  
    cout << maxOperations(X, Y);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG 
{
  
// Function to return the maximum operations
// required to convert X to Y
static int maxOperations(int X, int Y)
{
  
    // X cannot be converted to Y
    if (X > Y)
        return -1;
  
    int diff = Y - X;
  
    // If the differecne is 1
    if (diff == 1)
        return -1;
  
    // If the difference is even
    if (diff % 2 == 0)
        return (diff / 2);
  
    // Add 3 to X and the new
    // difference will be even
    return (1 + ((diff - 3) / 2));
}
  
// Driver code
public static void main(String []args) 
{
    int X = 5, Y = 16;
  
    System.out.println(maxOperations(X, Y));
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach 
  
# Function to return the maximum operations 
# required to convert X to Y 
def maxOperations(X, Y) : 
  
    # X cannot be converted to Y 
    if (X > Y) :
        return -1; 
  
    diff = Y - X; 
  
    # If the differecne is 1 
    if (diff == 1) :
        return -1; 
  
    # If the difference is even 
    if (diff % 2 == 0) :
        return (diff // 2); 
  
    # Add 3 to X and the new 
    # difference will be even 
    return (1 + ((diff - 3) // 2)); 
  
# Driver code 
if __name__ == "__main__" : 
  
    X = 5; Y = 16; 
  
    print(maxOperations(X, Y)); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;                    
  
class GFG
{
   
// Function to return the maximum operations
// required to convert X to Y
static int maxOperations(int X, int Y)
{
   
    // X cannot be converted to Y
    if (X > Y)
        return -1;
   
    int diff = Y - X;
   
    // If the differecne is 1
    if (diff == 1)
        return -1;
   
    // If the difference is even
    if (diff % 2 == 0)
        return (diff / 2);
   
    // Add 3 to X and the new
    // difference will be even
    return (1 + ((diff - 3) / 2));
}
   
// Driver code
public static void Main(String []args) 
{
    int X = 5, Y = 16;
   
    Console.WriteLine(maxOperations(X, Y));
}
}
  
// This code is contributed by PrinciRaj1992

输出：

5

时间复杂度： O(1)