仅通过加号运动检查是否可以精确地Z步从原点到达(x，y)

给定点(x，y) ，任务是检查是否有可能以正Z步从原点到达(x，y) 。从给定点(x，y)，我们只能在四个方向上向左(x – 1，y) ，向右(x + 1，y) ，向上(x，y + 1)和向下(x，y – 1)移动) 。
例子：

Input: x = 5, y = 5, z = 11 
Output: Not Possible
Input: x = 10, y = 15, z = 25 
Output: Possible 
 

方法：

1. 从原点到(x，y)的最短路径是| x | + | y | 。
2. 因此，很明显，如果Z小于| x | + | y | ，那么我们就无法精确地在Z步中从原点到达(x，y) 。
3. 如果步数不小于| x | + | y |然后，我们必须检查以下两个条件，以检查是否可以达到(x，y) ：
   • 如果Z≥| x | + | y | ， 和
   • 如果(Z – | x | + | y |)％2为0 。
4. 对于上述步骤中的第二个条件，如果达到(x，y) ，则可以再执行两个步骤，例如(x，y)–>(x，y + 1)–>(x，y)到相同位置(x，y) 。而且只有当它们之间的差异相等时才有可能。

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 it is possible to
// reach (x, y) from origin in exactly z steps
void possibleToReach(int x, int y, int z)
{
 
    // Condition if we can't reach in Z steps
    if (z < abs(x) + abs(y)
        || (z - abs(x) - abs(y)) % 2) {
        cout << "Not Possible" << endl;
    }
    else
        cout << "Possible" << endl;
}
 
// Driver Code
int main()
{
    // Destination point coordinate
    int x = 5, y = 5;
 
    // Number of steps allowed
    int z = 11;
 
    // Function Call
    possibleToReach(x, y, z);
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to check if it is possible
// to reach (x, y) from origin in
// exactly z steps
static void possibleToReach(int x, int y, int z)
{
     
    // Condition if we can't reach in Z steps
    if (z < Math.abs(x) + Math.abs(y) ||
       (z - Math.abs(x) - Math.abs(y)) % 2 == 1)
    {
        System.out.print("Not Possible" + "\n");
    }
    else
        System.out.print("Possible" + "\n");
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Destination point coordinate
    int x = 5, y = 5;
 
    // Number of steps allowed
    int z = 11;
 
    // Function Call
    possibleToReach(x, y, z);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
 
# Function to check if it is possible to
# reach (x, y) from origin in exactly z steps
def possibleToReach(x, y, z):
 
    #Condition if we can't reach in Z steps
    if (z < abs(x) + abs(y) or
       (z - abs(x) - abs(y)) % 2):
        print("Not Possible")
    else:
        print("Possible")
 
# Driver Code
if __name__ == '__main__':
     
    # Destination pocoordinate
    x = 5
    y = 5
 
    # Number of steps allowed
    z = 11
 
    # Function call
    possibleToReach(x, y, z)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to check if it is possible
// to reach (x, y) from origin in
// exactly z steps
static void possibleToReach(int x, int y, int z)
{
     
    // Condition if we can't reach in Z steps
    if (z < Math.Abs(x) + Math.Abs(y) ||
       (z - Math.Abs(x) - Math.Abs(y)) % 2 == 1)
    {
        Console.Write("Not Possible" + "\n");
    }
    else
        Console.Write("Possible" + "\n");
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Destination point coordinate
    int x = 5, y = 5;
 
    // Number of steps allowed
    int z = 11;
 
    // Function Call
    possibleToReach(x, y, z);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

// javascript program for the above approach
 
 
// Function to check if it is possible
// to reach (x, y) from origin in
// exactly z steps
function possibleToReach(x , y , z)
{
     
    // Condition if we can't reach in Z steps
    if (z < Math.abs(x) + Math.abs(y) ||
       (z - Math.abs(x) - Math.abs(y)) % 2 == 1)
    {
        document.write("Not Possible" + "\n");
    }
    else
        document.write("Possible" + "\n");
}
 
// Driver Code
// Destination povar coordinate
var x = 5, y = 5;
 
// Number of steps allowed
var z = 11;
 
// Function Call
possibleToReach(x, y, z);
 
// This code is contributed by Amit Katiyar

输出：

Not Possible

时间复杂度： O(1)
辅助空间： O(1)