检查 X 是否可以通过在每个操作中转换为 3 * (X / 2) 或 X – 1 来转换为 Y

给定两个正整数X和Y ，任务是通过重复将X的值更改为(3 * X / 2) （如果X是偶数）或(X – 1)来检查X 是否可以转换为Y。如果可以将X转换为Y ，则打印“是” 。否则，打印“否” 。

例子：

Input: X = 6, Y = 8
Output: Yes
Explanation:
Operation 1: Convert X(= 6) to 3*X/2 ( = 3 * (6 / 2) = 9).
Operation 2: Convert X(= 9) to (X – 1) (= 8).
Therefore, the total number of operations required is 2.

Input: X = 3, Y = 6
Output: No

方法：根据以下观察可以解决给定的问题：

• 如果X的值至少是Y ，那么X总是可以使用第二个操作转换为Y ，即，将X减少1 。
• 如果X是偶数，那么它可以转换为(3 * (X / 2)) ，对于大于0 的所有偶数，它都大于X 。
• 如果X的值是奇数，那么X可以转换为(X – 1) ，可以转换为(3 * (X – 1)/2) ，它大于X 。
• 因此，从上述观察，X转化成Y是总是可能的X> 3。
• 需要考虑以下基本情况：
  • X = 1：仅当Y = 1时才可能进行转换。
  • X = 2 或 X = 3：仅当Y ≤ 3时才可能进行转换。
  • 在所有其他情况下，转换是不可能的。

请按照以下步骤解决问题：

• 如果值X大于4 ，则打印“是” 。
• 如果值X为1且Y为1 ，则打印“是” 。
• 如果值X是2或3并且Y小于4 ，则打印“是” 。
• 否则，为所有其他情况打印“否” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if X can be
// made equal to Y by converting
// X to (3*X/2) or (X - 1)
void check(int X, int Y)
{
    // Conditions for possible conversion
    if (X > 3) {
        cout << "Yes";
    }
 
    else if (X == 1 and Y == 1) {
        cout << "Yes";
    }
 
    else if (X == 2 and Y <= 3) {
        cout << "Yes";
    }
 
    else if (X == 3 and Y <= 3) {
        cout << "Yes";
    }
 
    // Otherwise, conversion
    // is not possible
    else {
        cout << "No";
    }
}
 
// Driver Code
int main()
{
    int X = 6, Y = 8;
    check(X, Y);
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
     
// Function to check if X can be
// made equal to Y by converting
// X to (3*X/2) or (X - 1)
static void check(int X, int Y)
{
     
    // Conditions for possible conversion
    if (X > 3)
    {
        System.out.print("Yes");
    }
 
    else if (X == 1 && Y == 1)
    {
        System.out.print("Yes");
    }
 
    else if (X == 2 && Y <= 3)
    {
        System.out.print("Yes");
    }
 
    else if (X == 3 && Y <= 3)
    {
        System.out.print("Yes");
    }
 
    // Otherwise, conversion
    // is not possible
    else
    {
        System.out.print("No");
    }
}
 
// Driver Code
public static void main (String[] args)
{
    int X = 6, Y = 8;
     
    check(X, Y);
}
}
 
// This code is contributed by AnkThon

Python3

# Python3 program for the above approach
 
# Function to check if X can be
# made equal to Y by converting
# X to (3*X/2) or (X - 1)
def check(X, Y):
     
    # Conditions for possible conversion
    if (X > 3):
        print("Yes")
 
    elif (X == 1 and Y == 1):
        print("Yes")
 
    elif (X == 2 and Y <= 3):
        print("Yes")
 
    elif (X == 3 and Y <= 3):
        print("Yes")
 
    # Otherwise, conversion
    # is not possible
    else:
        print("No")
 
# Driver Code
if __name__ == '__main__':
     
    X = 6
    Y = 8
     
    check(X, Y)
 
# This code is contributed by ipg2016107

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if X can be
// made equal to Y by converting
// X to (3*X/2) or (X - 1)
static void check(int X, int Y)
{
     
    // Conditions for possible conversion
    if (X > 3)
    {
        Console.WriteLine("Yes");
    }
 
    else if (X == 1 && Y == 1)
    {
        Console.WriteLine("Yes");
    }
 
    else if (X == 2 && Y <= 3)
    {
        Console.WriteLine("Yes");
    }
 
    else if (X == 3 && Y <= 3)
    {
        Console.WriteLine("Yes");
    }
 
    // Otherwise, conversion
    // is not possible
    else
    {
        Console.WriteLine("No");
    }
}
 
// Driver Code
public static void Main(string[] args)
{
    int X = 6, Y = 8;
     
    check(X, Y);
}
}
 
// This code is contributed by avijitmondal1998

Javascript

 
    // JavaScript program for the above approach
     
    // Function to check if X can be
    // made equal to Y by converting
    // X to (3*X/2) or (X - 1)
    function check(X, Y)
    {
 
        // Conditions for possible conversion
        if (X > 3)
        {
            document.write("Yes");
        }
 
        else if (X == 1 && Y == 1)
        {
            document.write("Yes");
        }
 
        else if (X == 2 && Y <= 3)
        {
            document.write("Yes");
        }
 
        else if (X == 3 && Y <= 3)
        {
            document.write("Yes");
        }
 
        // Otherwise, conversion
        // is not possible
        else
        {
            document.write("No");
        }
    }
     
    let X = 6, Y = 8;
      
    check(X, Y);
 

输出：

Yes

时间复杂度： O(1)
辅助空间： O(1)