通过给定步骤检查是否可以从(1，0)到达(X，Y)

给定两个正整数X和Y ，任务是检查是否可以通过给定的步骤从(1，0)达到(X，Y) 。在每个步骤中，来自任何单元格(a，b)的可能移动是(a，b + a)或(a + b，b) 。如果可能，打印“是” 。否则，打印“否” 。

例子：

Input: X = 2, Y = 7
Output: Yes
Explanation: Sequence of moves to reach (2, 7) are: (1, 0) -> (1, 1) -> (2, 1) -> (2, 3) -> (2, 5) -> (2, 7).

Input: X = 30, Y = 24
Output: No

天真的方法：最简单的方法是尝试使用操作(X – Y，Y)或(X，Y – X)递归地从点(X，Y)移至(1，0) ，直到等于( 1，0) 。如果X坐标小于1或Y坐标小于0 ，则不可能达到(1，0) 。因此，打印“否” 。否则，如果达到(1，0) ，则打印“是” 。

时间复杂度： O(2 ^{log(min(X，Y))} )
辅助空间： O(1)

高效方法：这个想法是观察以下属性：

• 尝试以相反的顺序解决问题，即可以通过隐性地移动到点(X-Y，Y)或(X，Y-X)从(X，Y)移至(1，0 ) 。
• 上面的属性可以表示为：

GCD(X, Y) = GCD(X, Y – X) or GCD(X – Y, Y)
where,  
Base Case is GCD(X, 0) = X
Now, notice that gcd of 1 and 0 i.e., gcd(1, 0) is 1.
Therefore, gcd of X and Y must also be 1 to reach (1, 0).

因此，根据以上观察，如果GCD(X，Y)= 1，则始终存在从(1，0)到(X，Y)的路径。如果给定的两个数字的GCD为1，则打印“是”。否则，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 find the GCD of two
// numbers a and b
int GCD(int a, int b)
{
    // Base Case
    if (b == 0)
        return a;
 
    // Recursively find the GCD
    else
        return GCD(b, a % b);
}
 
// Function to check if (x, y) can be
// reached from (1, 0) from given moves
void check(int x, int y)
{
    // If GCD is 1, then print "Yes"
    if (GCD(x, y) == 1) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
}
 
// Driver Code
int main()
{
    // Given X and Y
    int X = 2, Y = 7;
 
    // Function call
    check(X, Y);
 
    return 0;
}

Java

// Java program for the
// above approach
import java.util.*;
class GFG{
 
// Function to find the
// GCD of two numbers a
// and b
static int GCD(int a,
               int b)
{
  // Base Case
  if (b == 0)
    return a;
 
  // Recursively find
  // the GCD
  else
    return GCD(b, a % b);
}
 
// Function to check if
// (x, y) can be reached
// from (1, 0) from given
// moves
static void check(int x,
                  int y)
{
  // If GCD is 1, then
  // print "Yes"
  if (GCD(x, y) == 1)
  {
    System.out.print("Yes");
  }
  else
  {
    System.out.print("No");
  }
}
 
// Driver Code
public static void main(String[] args)
{
  // Given X and Y
  int X = 2, Y = 7;
 
  // Function call
  check(X, Y);
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 program for the above approach
 
# Function to find the GCD of two
# numbers a and b
def GCD(a, b):
     
    # Base Case
    if (b == 0):
        return a
         
    # Recursively find the GCD
    else:
        return GCD(b, a % b)
 
# Function to check if (x, y) can be
# reached from (1, 0) from given moves
def check(x, y):
     
    # If GCD is 1, then pr"Yes"
    if (GCD(x, y) == 1):
        print("Yes")
    else:
        print("No")
 
# Driver Code
if __name__ == '__main__':
     
    # Given X and Y
    X = 2
    Y = 7
 
    # Function call
    check(X, Y)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the
// above approach
using System;
 
class GFG{
   
// Function to find the
// GCD of two numbers a
// and b
static int GCD(int a, int b)
{
   
  // Base Case
  if (b == 0)
    return a;
 
  // Recursively find
  // the GCD
  else
    return GCD(b, a % b);
}
 
// Function to check if
// (x, y) can be reached
// from (1, 0) from given
// moves
static void check(int x, int y)
{
   
  // If GCD is 1, then
  // print "Yes"
  if (GCD(x, y) == 1)
  {
    Console.WriteLine("Yes");
  }
  else
  {
    Console.WriteLine("No");
  }
}
 
// Driver Code
public static void Main()
{
   
  // Given X and Y
  int X = 2, Y = 7;
 
  // Function call
  check(X, Y);
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript

// JavaScript program for the above approach
 
// Function to find the GCD of two
// numbers a and b
function GCD(a, b)
{
 
    // Base Case
    if (b == 0)
        return a;
 
    // Recursively find the GCD
    else
        return GCD(b, a % b);
}
 
// Function to check if (x, y) can be
// reached from (1, 0) from given moves
function check(x, y)
{
 
    // If GCD is 1, then print "Yes"
    if (GCD(x, y) == 1) {
        document.write("Yes");
    }
    else {
        document.write("No");
    }
}
 
// Driver Code
 
    // Given X and Y
    let X = 2, Y = 7;
 
    // Function call
    check(X, Y);
 
// This code is contributed by Surbhi Tyagi.

输出：

Yes

时间复杂度： O(log(min(X，Y))
辅助空间： O(1)