检查是否可以在 N 步中从 (0, 0) 移动到 (x, y)

给定一个点 (x, y)。找出是否有可能以恰好 n 步从 (0, 0) 移动到 (x, y)。 4 种类型的步骤是有效的，您可以从 (a, b) 点移动到 (a, b+1)、(a, b-1)、(a-1, b)、(a+1, b)
例子：

Input: x = 0, y = 0, n = 2
Output: POSSIBLE

Input: x = 1, y = 1, n = 3 
Output: IMPOSSIBLE

方法：

In the shortest path, one can move from (0, 0) to (x, y) in |x| + |y|. So, it is not possible to move from (0, 0) to (x, y) in less than |x| + |y| steps. After reaching one can take two more steps as (x, y) -> (x, y+1) -> (x, y). 
 

所以，如果

n >= |x| + |y| and ( n-( |x| + |y| ) ) % 2 = 0. 
 

下面是上述方法的实现：

C++

// CPP program to check whether it is possible
// or not to move from (0, 0) to (x, y)
// in exactly n steps
#include 
using namespace std;
 
// Function to check whether it is possible
// or not to move from (0, 0) to (x, y)
// in exactly n steps
bool Arrive(int a, int b, int n)
{
    if (n >= abs(a) + abs(b) and (n - (abs(a) + abs(b))) % 2 == 0)
        return true;
 
    return false;
}
 
// Driver code
int main()
{
    int a = 5, b = 5, n = 11;
 
    if (Arrive(a, b, n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java program to check whether it is possible
// or not to move from (0, 0) to (x, y)
// in exactly n steps
import java.io.*;
 
public class GFG {
 
// Function to check whether it is possible
// or not to move from (0, 0) to (x, y)
// in exactly n steps
static boolean Arrive(int a, int b, int n)
{
    if (n >= Math.abs(a) + Math.abs(b) && (n - (Math.abs(a) + Math.abs(b))) % 2 == 0)
        return true;
 
    return false;
}
 
// Driver code
int main()
{
 
    return 0;
}
 
    public static void main (String[] args) {
         
    int a = 5, b = 5, n = 11;
 
    if (Arrive(a, b, n))
        System.out.println( "Yes");
    else
        System.out.println( "No");
    }
}
//This code is contributed by shs..

Python3

# Python3 program to check whether
# it is possible or not to move from
# (0, 0) to (x, y) in exactly n steps
 
# Function to check whether it is
# possible or not to move from
# (0, 0) to (x, y) in exactly n steps
def Arrive(a, b, n):
     
    if (n >= abs(a) + abs(b) and
       (n - (abs(a) + abs(b))) % 2 == 0):
        return True
 
    return False
 
# Driver code
a = 5
b = 5
n = 11
 
if (Arrive(a, b, n)):
    print("Yes")
else:
    print("No")
 
# This code is contributed
# by Yatin Gupta

C#

// C# program to check whether
// it is possible or not to
// move from (0, 0) to (x, y)
// in exactly n steps
using System;
 
class GFG
{
 
// Function to check whether it
// is possible or not to move
// from (0, 0) to (x, y) in
// exactly n steps
static bool Arrive(int a, int b, int n)
{
    if (n >= Math.Abs(a) + Math.Abs(b) &&
       (n - (Math.Abs(a) + Math.Abs(b))) % 2 == 0)
        return true;
 
    return false;
}
 
// Driver code
public static void Main ()
{
    int a = 5, b = 5, n = 11;
 
    if (Arrive(a, b, n))
        Console.WriteLine( "Yes");
    else
        Console.WriteLine( "No");
    }
}
 
// This code is contributed by shashank

PHP

= abs($a) + abs($b) and
       ($n - (abs($a) + abs($b))) % 2 == 0)
        return true;
 
    return false;
}
 
// Driver code
$a = 5; $b = 5; $n = 11;
 
if (Arrive($a, $b, $n))
    echo "Yes";
else
    echo "No";
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

Javascript

 
// JavaScript program to check whether it is
// possible or not to move from (0, 0) to (x, y)
// in exactly n steps
 
// Function to check whether it is possible
// or not to move from (0, 0) to (x, y)
// in exactly n steps
function Arrive(a, b, n)
{
    if (n >= Math.abs(a) + Math.abs(b) &&
       (n - (Math.abs(a) + Math.abs(b))) % 2 == 0)
        return true;
 
    return false;
}
 
// Driver code
var a = 5, b = 5, n = 11;
 
if (Arrive(a, b, n))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by Ankita saini
 

输出：

No