通过重复将 X 乘以 2 或在末尾附加 1 将 X 转换为 Y

给定两个正整数X和Y ，任务是检查是否可以通过将X乘以2或在X的末尾附加1将数字X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 = 100, Y = 40021
Output: Yes
Explanation:
Below are the operations performed to convert X into Y:
Operation 1: Multiply X(= 100) by 2, modifies the value X to 200.
Operation 2: Append 1 at the end of X(= 200), modifies the value X to 2001.
Operation 3: Multiply X(= 2001) by 2, modifies the value X to 4002.
Operation 4: Append 1 at the end of X(= 4002), modifies the value X to 40021.
Therefore, from the above operations, it can be seen that the value X can be converted into Y. Hence, print Yes.

Input: X = 17 and Y = 35
Output: No

方法：给定的问题可以通过以相反的方式执行操作来解决，即尝试将值Y转换为X 。请按照以下步骤解决问题：

• 迭代直到Y的值大于X并执行以下步骤：
  • 如果Y的最后一位的值为 1，则将Y的值除以10 。
  • 否则，如果Y的值可被 2 整除，则将Y除以2 。
  • 否则，跳出循环。
• 完成上述步骤后，如果Y的值与X的值相同，则打印Yes 。否则，打印No 。

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
// converted to Y by multiplying
// X by 2 or appending 1 at the end
void convertXintoY(int X, int Y)
{
    // Iterate until Y is at least X
    while (Y > X) {
  
        // If Y is even
        if (Y % 2 == 0)
            Y /= 2;
  
        // If the last digit of Y is 1
        else if (Y % 10 == 1)
            Y /= 10;
  
        // Otherwise
        else
            break;
    }
  
    // Check if X is equal to Y
    if (X == Y)
        cout << "Yes";
    else
        cout << "No";
}
  
// Driver Code
int main()
{
    int X = 100, Y = 40021;
    convertXintoY(X, Y);
  
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to check if X can be
// converted to Y by multiplying
// X by 2 or appending 1 at the end
static void convertXintoY(int X, int Y)
{
    // Iterate until Y is at least X
    while (Y > X) {
  
        // If Y is even
        if (Y % 2 == 0)
            Y /= 2;
  
        // If the last digit of Y is 1
        else if (Y % 10 == 1)
            Y /= 10;
  
        // Otherwise
        else
            break;
    }
  
    // Check if X is equal to Y
    if (X == Y)
        System.out.print("Yes");
    else
        System.out.print("No");
}
  
// Driver Code
public static void main(String[] args)
{
    int X = 100, Y = 40021;
    convertXintoY(X, Y);
}
}
  
// This code is contributed by sanjoy_62.

Python3

# Python program for the above approach
  
# Function to check if X can be
# converted to Y by multiplying
# X by 2 or appending 1 at the end
def convertXintoY(X, Y):
    # Iterate until Y is at least X
    while (Y > X):
  
        # If Y is even
        if (Y % 2 == 0):
            Y //= 2
  
        # If the last digit of Y is 1
        elif (Y % 10 == 1):
            Y //= 10
  
        # Otherwise
        else:
            break
  
    # Check if X is equal to Y
    if (X == Y):
        print("Yes")
    else:
        print("No")
  
# Driver Code
if __name__ == '__main__':
    X,Y = 100, 40021
    convertXintoY(X, Y)
  
# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
  
class GFG{
  
// Function to check if X can be
// converted to Y by multiplying
// X by 2 or appending 1 at the end
static void convertXintoY(int X, int Y)
{
      
    // Iterate until Y is at least X
    while (Y > X) 
    {
          
        // If Y is even
        if (Y % 2 == 0)
            Y /= 2;
  
        // If the last digit of Y is 1
        else if (Y % 10 == 1)
            Y /= 10;
  
        // Otherwise
        else
            break;
    }
  
    // Check if X is equal to Y
    if (X == Y)
        Console.Write("Yes");
    else
        Console.Write("No");
}
  
// Driver Code
public static void Main(String[] args)
{
    int X = 100, Y = 40021;
      
    convertXintoY(X, Y);
}
}
  
// This code is contributed by shivanisinghss2110

Javascript

// JavaScript program for the above approach
  
        // Function to check if X can be
        // converted to Y by multiplying
        // X by 2 or appending 1 at the end
        function convertXintoY(X, Y)
        {
            // Iterate until Y is at least X
            while (Y > X) {
  
                // If Y is even
                if (Y % 2 == 0)
                    Y = parseInt(Y / 2);
  
                // If the last digit of Y is 1
                else if (Y % 10 == 1)
                    Y = parseInt(Y /= 10);
  
                // Otherwise
                else
                    break;
            }
  
            // Check if X is equal to Y
            if (X == Y)
                document.write("Yes");
            else
                document.write("No");
        }
  
        // Driver Code
        let X = 100, Y = 40021;
        convertXintoY(X, Y);
  
  // This code is contributed by Potta Lokesh
    

输出：

Yes

时间复杂度： log(Y)
辅助空间： O(1)