检查点 (X, Y) 是否可以从原点 (0, 0) 到达同时垂直跳跃 1 和 N
给定一个正整数N和坐标(X, Y) ，任务是检查是否有可能从(0, 0)到达(X, Y)并同时在垂直方向上跳跃1和N。如果可以到达(X, Y)则打印Yes 。否则，打印No 。
例子：
Input: N = 2, X = 5, Y = 4
Output: Yes
Explanation:
Following are the possible moves that leads to point (X, Y) from the current coordinate (0, 0):
- Performing the jump of (2, 1) from the current coordinate modifies it to (2, 1).
- Performing the jump of (2, 1) from the current coordinate modifies it to (4, 2).
- Performing the jump of (1, 2) from the current coordinate modifies it to (5, 4).
Input: N = 3, X = 5, Y = 4
Output: No
方法：给定的问题可以基于观察到有 4 种可能的方式可以从任何坐标跳转来解决，它们是(1, N)、(1, -N)、(-1, N)和(-1, -N) 。此外，问题可以分为3种不同的情况：
- 情况 1 - 其中 N 是偶数：在这种情况下，任何坐标 (X, Y) 都是可达的。让我们一次考虑 X 和 Y。对于坐标 X，可以遵循(1, N), (1, -N), (1, N), (1, -N), …的顺序跳转。结果坐标将是(X, 0) （即所需坐标）或(X, N) 。由于N是偶数，因此也可以遵循跳转序列(N, -1), (-N, -1), (N, -1), (N, -1), ...到达(X, 0) .同理， (0,Y)可以到达，结合跳跃的顺序到达(X,0)和(0,Y)，(X,Y)就可以到达。
- 情况 2 – 其中 N 为奇数，X 和 Y 均为偶数或奇数：这些情况可以通过遵循类似于情况 1的一系列操作来处理。
- 情况 3 – 其中 N 为奇数，X 和 Y 具有不同的奇偶性：在这种情况下，没有可能的跳转序列到达(X, Y)，因为根据情况 2， (X + 1, Y)和(X, Y + 1)必须是可达的，因为它们都将具有相同的奇偶校验，并且没有可能的操作序列来覆盖(1, 0)或(0, 1)的距离。
因此，唯一不可能从(0, 0)到达(X, Y)的情况是N为奇数且X和Y具有不同的奇偶性，即X为偶数且Y为奇数，反之亦然。
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, Y) is reachable
// from (0, 0) using the jumps of given type
string checkReachability(int N, int X, int Y)
{
// Case where source & destination
// are the same
if (X == 0 && Y == 0) {
return "YES";
}
// Check for even N (X, Y) is
// reachable or not
if (N % 2 == 0) {
return "YES";
}
// If N is odd and parity of X and
// Y is different return, no valid
// sequence of jumps exist
else {
if (X % 2 != Y % 2) {
return "NO";
}
else {
return "YES";
}
}
}
// Driver Code
int main()
{
int N = 2;
int X = 5, Y = 4;
cout << checkReachability(N, X, Y);
return 0;
} Java
// Java code for the above approach
import java.io.*;
class GFG
{
// Function to check if (X, Y) is reachable
// from (0, 0) using the jumps of given type
static String checkReachability(int N, int X, int Y)
{
// Case where source & destination
// are the same
if (X == 0 && Y == 0) {
return "YES";
}
// Check for even N (X, Y) is
// reachable or not
if (N % 2 == 0) {
return "YES";
}
// If N is odd and parity of X and
// Y is different return, no valid
// sequence of jumps exist
else {
if (X % 2 != Y % 2) {
return "NO";
}
else {
return "YES";
}
}
}
// Driver Code
public static void main (String[] args) {
int N = 2;
int X = 5, Y = 4;
System.out.println(checkReachability(N, X, Y));
}
}
// This code is contributed by Potta LokeshPython3
# Python 3 program for the above approach
# Function to check if (X, Y) is reachable
# from (0, 0) using the jumps of given type
def checkReachability(N, X, Y):
# Case where source & destination
# are the same
if (X == 0 and Y == 0):
return "YES"
# Check for even N (X, Y) is
# reachable or not
if (N % 2 == 0):
return "YES"
# If N is odd and parity of X and
# Y is different return, no valid
# sequence of jumps exist
else:
if (X % 2 != Y % 2):
return "NO"
else:
return "YES"
# Driver Code
if __name__ == '__main__':
N = 2
X = 5
Y = 4
print(checkReachability(N, X, Y))
# This code is contributed by SURENDRA_GANGWAR.C#
// C# code for the above approach
using System;
class GFG
{
// Function to check if (X, Y) is reachable
// from (0, 0) using the jumps of given type
static string checkReachability(int N, int X, int Y)
{
// Case where source & destination
// are the same
if (X == 0 && Y == 0) {
return "YES";
}
// Check for even N (X, Y) is
// reachable or not
if (N % 2 == 0) {
return "YES";
}
// If N is odd and parity of X and
// Y is different return, no valid
// sequence of jumps exist
else {
if (X % 2 != Y % 2) {
return "NO";
}
else {
return "YES";
}
}
}
// Driver Code
public static void Main (string[] args) {
int N = 2;
int X = 5, Y = 4;
Console.WriteLine(checkReachability(N, X, Y));
}
}
// This code is contributed by AnkThonJavascript
输出：
YES时间复杂度： O(1)
辅助空间： O(1)