确定给定线将通过的单位面积的平方数。

给定一条线的 2 个端点(x1, y1)和(x2, y2) ，任务是确定该线将通过的单位面积的方格数。

例子：

Input: (x1 = 1, y1 = 1), (x2 = 4, y2 = 3) 
Output: 4 
In the diagram above the line is passing through 4 squares
Input: (x1 = 0, y1 = 0), (x2 = 2, y2 = 2) 
Output: 2 
 

方法：
让，

Dx = (x2 - x1)
Dy = (y2 - y1)

所以，

x = x1 + Dx * t
y = y1 + Dy * t

我们必须在 (0, 1] 中为 t 找到 (x, y)。
因为 x 和 y 是整数，所以 Dx 和 Dy 必须能被 t 整除。此外，t 不能是无理数，因为 Dx 和 Dy 是整数。
因此让t = p / q 。
Dx 和 Dy 必须能被 q 整除。所以 Dx 和 Dy 的 GCD 一定是 q。
或者，q = GCD(Dx, Dy)。
只有 GCD(Dx, Dy) 最小的子问题。
下面是上述方法的实现：

C++

#include
using namespace std;
 
// Function to return the required position
int noOfSquares(int x1, int y1, int x2, int y2)
{
    int dx = abs(x2 - x1);
    int dy = abs(y2 - y1);
 
    int ans = dx + dy - __gcd(dx, dy);
 
    cout<

---

Java

// Java program to determine the number
// of squares that line will pass through
class GFG
{
static int __gcd(int a, int b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);
     
}
 
// Function to return the required position
static void noOfSquares(int x1, int y1,
                        int x2, int y2)
{
    int dx = Math.abs(x2 - x1);
    int dy = Math.abs(y2 - y1);
 
    int ans = dx + dy - __gcd(dx, dy);
 
    System.out.println(ans);
}
 
// Driver Code
public static void main(String []args)
{
    int x1 = 1, y1 = 1, x2 = 4, y2 = 3;
 
    noOfSquares(x1, y1, x2, y2);
}
}
 
// This code contributed by Rajput-Ji

Python3

# Python3 program to determine the number
# of squares that line will pass through
from math import gcd
 
# Function to return the required position
def noOfSquares(x1, y1, x2, y2) :
 
    dx = abs(x2 - x1);
    dy = abs(y2 - y1);
 
    ans = dx + dy - gcd(dx, dy);
 
    print(ans);
 
# Driver Code
if __name__ == "__main__" :
 
    x1 = 1; y1 = 1; x2 = 4; y2 = 3;
 
    noOfSquares(x1, y1, x2, y2);
 
# This code is contributed by Ryuga

C#

using System;
 
class GFG
{
static int __gcd(int a, int b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);
     
}
 
// Function to return the required position
static void noOfSquares(int x1, int y1,
                        int x2, int y2)
{
    int dx = Math.Abs(x2 - x1);
    int dy = Math.Abs(y2 - y1);
 
    int ans = dx + dy - __gcd(dx, dy);
 
    Console.WriteLine(ans);
}
 
// Driver Code
static void Main()
{
    int x1 = 1, y1 = 1, x2 = 4, y2 = 3;
 
    noOfSquares(x1, y1, x2, y2);
}
}
 
// This code is contributed by mits

Javascript

 
function __gcd(a, b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);
     
}
 
// Function to return the required position
function noOfSquares(x1, y1, x2, y2)
{
    var dx = Math.abs(x2 - x1);
    var dy = Math.abs(y2 - y1);
 
    var ans = dx + dy - __gcd(dx, dy);
 
    document.write(ans);
}
 
// Driver Code
var x1 = 1, y1 = 1, x2 = 4, y2 = 3;
noOfSquares(x1, y1, x2, y2);
 
// This code is contributed by noob2000.

输出：

4

时间复杂度： O(1)