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. 将任何质数添加到X 。
2. 从Y 中减去任何素数。

如果无法将X转换为Y ，则打印所需的此类操作的最大数量或-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：由于任务是最大化操作，所以必须在每次操作中将最小可能值添加到X。由于该值必须是质数，因此可以使用最小的两个质数，即2和3，因为它们都是质数并且可以涵盖偶数和奇数奇偶校验。现在分三种情况：

• 如果X > Y则答案将为-1，因为使用给定的操作无法使X等于Y。
• 如果X = Y那么答案将是0 。
• 如果X < Y则计算P = Y – X并且，
  • 如果P = 1，那么答案将是-1，因为1不是素数，不能加减。
  • 如果P是偶数，那么2可以重复添加到 X 中，答案将是P / 2
  • 如果P是偶数，则将3添加到X然后可以再次将2重复添加到新的X以使其等于Y ，在这种情况下，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 difference 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 difference 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 difference 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 difference 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

Javascript

 
// JavaScript implementation of the approach
 
// Function to return the maximum operations
// required to convert X to Y
function maxOperations(X, Y)
{
 
    // X cannot be converted to Y
    if (X > Y)
        return -1;
 
    let diff = Y - X;
 
    // If the difference 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
 
    let X = 5, Y = 16;
 
    document.write(maxOperations(X, Y));
 
// This code is contributed by _saurabh_jaiswal   
 

输出：

5

时间复杂度： O(1)

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