检查是否可以精确地以K步从(0，0)移至(X，Y)

给定二维平面中的点(X，Y)和整数K ，任务是检查是否有可能以正好K的运动从(0，0)移至给定的点(X，Y) 。单步移动可从(X，Y)到达的位置是(X，Y + 1) ， (X，Y – 1) ， (X + 1，Y)和(X – 1，Y) 。
例子：

Input: X = 0, Y = 0, K = 2 
Output: Yes 
Move 1: (0, 0) -> (0, 1) 
Move 2: (0, 1) -> (0, 0)
Input: X = 5, Y = 8, K = 20 
Output: No 
 

方法：显然，从(0，0 )到达(X，Y)的最短路径将是minMoves =(| X | + | Y |) 。因此，如果K ，则不可能达到(X，Y)，但是如果K≥minMoves，则在达到(X，Y)后，必须达到minMoves的移动数量，其余(K – minMoves)的移动数量必须等于以便在接下来的动作中保持在这一点上。
因此，只有当K≥(| X | + | Y |)和(K –(| X | + | Y |))％2 = 0时，才可能从(0，0 )达到(X，Y) 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function that returns true if it is
// possible to move from (0, 0) to
// (x, y) in exactly k moves
bool isPossible(int x, int y, int k)
{
    // Minimum moves required
    int minMoves = abs(x) + abs(y);
 
    // If possible
    if (k >= minMoves && (k - minMoves) % 2 == 0)
        return true;
 
    return false;
}
 
// Driver code
int main()
{
    int x = 5, y = 8, k = 20;
 
    if (isPossible(x, y, k))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
    // Function that returns true if it is
    // possible to move from (0, 0) to
    // (x, y) in exactly k moves
    static boolean isPossible(int x, int y, int k)
    {
        // Minimum moves required
        int minMoves = Math.abs(x) + Math.abs(y);
     
        // If possible
        if (k >= minMoves && (k - minMoves) % 2 == 0)
            return true;
     
        return false;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int x = 5, y = 8, k = 20;
     
        if (isPossible(x, y, k))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
 
# Function that returns true if it is
# possible to move from (0, 0) to
# (x, y) in exactly k moves
def isPossible(x, y, k):
     
    # Minimum moves required
    minMoves = abs(x) + abs(y)
 
    # If possible
    if (k >= minMoves and (k - minMoves) % 2 == 0):
        return True
 
    return False
 
# Driver code
x = 5
y = 8
k = 20
 
if (isPossible(x, y, k)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
class GFG
{
     
    // Function that returns true if it is
    // possible to move from (0, 0) to
    // (x, y) in exactly k moves
    static bool isPossible(int x, int y, int k)
    {
        // Minimum moves required
        int minMoves = Math.Abs(x) + Math.Abs(y);
     
        // If possible
        if (k >= minMoves && (k - minMoves) % 2 == 0)
            return true;
     
        return false;
    }
     
    // Driver code
    public static void Main ()
    {
        int x = 5, y = 8, k = 20;
     
        if (isPossible(x, y, k))
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}
 
// This code is contributed by Nidhi

Javascript

 
// javascript implementation of the approach
 
     
// Function that returns true if it is
// possible to move from (0, 0) to
// (x, y) in exactly k moves
function isPossible(x , y , k)
{
    // Minimum moves required
    var minMoves = Math.abs(x) + Math.abs(y);
 
    // If possible
    if (k >= minMoves && (k - minMoves) % 2 == 0)
        return true;
 
    return false;
}
 
// Driver code
 
var x = 5, y = 8, k = 20;
 
if (isPossible(x, y, k))
    document.write("Yes");
else
    document.write("No");
 
// This code contributed by shikhasingrajput
 

输出：

No