正方形内的积分坐标计数

给定正方形的左下角和右上角坐标(x1，y1)和(x2，y2) ，任务是计算严格位于正方形内的整数坐标的数量。
例子：

Input: x1 = 1, y1 = 1, x2 = 5, x3 = 5 
Output: 9 
Explanation: 
Below is the square for the given coordinates: 
 

Input: x1 = 1, y1 = 1, x2 = 4, x3 = 4 
Output: 4 
 

方法：给定正方形的右下角坐标和右上角坐标的x和y坐标之差分别给出了x坐标和y坐标在正方形相对两侧之间的数量积分点。严格位于正方形内的点总数由下式给出：

count = (x2 – x1 – 1) * (y2 – y1 – 1) 
 

例如：

In the above figure: 
1. The total number of integral points inside base of the square is (x2 – x1 – 1). 
2. The total number of integral points inside height of the square is (y2 – y1 – 1). 
These (x2 – x1 – 1) integrals points parallel to the base of the square repeats (y2 – y1 – 1) number of times. Therefore the total number of integral points is given by (x2 – x1 – 1)*(y2 – y1 – 1). 
 

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to calculate the integral
// points inside a square
void countIntgralPoints(int x1, int y1,
                        int x2, int y2)
{
    cout << (y2 - y1 - 1) * (x2 - x1 - 1);
}
 
// Driver Code
int main()
{
    int x1 = 1, y1 = 1;
    int x2 = 4, y2 = 4;
 
    countIntgralPoints(x1, y1, x2, y2);
    return 0;
}

Java

// Java program for the above approach
 
class GFG {
 
// Function to calculate the integral
// points inside a square
static void countIntgralPoints(int x1, int y1,
                               int x2, int y2)
{
    System.out.println((y2 - y1 - 1) *
                       (x2 - x1 - 1));
}
 
// Driver Code
public static void main(String args[])
{
    int x1 = 1, y1 = 1;
    int x2 = 4, y2 = 4;
     
    countIntgralPoints(x1, y1, x2, y2);
}
}
 
// This code is contributed by rutvik_56

Python3

# Python3 program for the above approach
 
# Function to calculate the integral
# points inside a square
def countIntgralPoints(x1, y1, x2, y2):
    print((y2 - y1 - 1) * (x2 - x1 - 1))
 
# Driver Code
if __name__ == '__main__':
 
    x1 = 1
    y1 = 1
    x2 = 4
    y2 = 4
 
    countIntgralPoints(x1, y1, x2, y2)
 
# This code is contributed by Samarth

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to calculate the integral
// points inside a square
static void countIntgralPoints(int x1, int y1,
                               int x2, int y2)
{
    Console.WriteLine((y2 - y1 - 1) *
                      (x2 - x1 - 1));
}
 
// Driver code
static void Main()
{
    int x1 = 1, y1 = 1;
    int x2 = 4, y2 = 4;
     
    countIntgralPoints(x1, y1, x2, y2);
}
}
 
// This code is contributed by divyeshrabadiya07

Javascript

 
// Javascript program for the above approach
 
// Function to calculate the integral
// points inside a square
function countIntgralPoints(x1, y1, x2, y2)
{
    document.write( (y2 - y1 - 1) * (x2 - x1 - 1));
}
 
// Driver Code
 
var x1 = 1, y1 = 1;
var x2 = 4, y2 = 4;
 
countIntgralPoints(x1, y1, x2, y2);
 

输出：

4

时间复杂度： O(1)