微客导航 » 文章资讯 » 应用上下测试来查找点相对于直线的位置的 C++ 程序

应用上下测试来查找点相对于直线的位置的 C++ 程序

2023-06-20 11:12:03 430

这是一个C++程序，用于应用上下测试来查找点相对于线的位置。对于平面上的任何点t(xt,yt)，它相对于连接m和n的线L的位置是通过计算标量s找到的-

Y = A xt + B yt + C

如果Y<0，t位于L的顺时针半平面内；如果Y>0，则t位于逆时针半平面上；如果Y=0，则t位于L上。

算法

Begin
   Take the points as input.
   For generating equation of the line, generate random numbers for coefficient of x and y (x1,x2,y1,y2) by using rand function at every time of compilation.
   Compute s as (y2 - y1) * x + (x1 - x2) * y + (x2 * y1 - x1 * y2).
   if (s < 0)
      Print "The point lies below the line or left side of the line".
   else if (s >0)
      print "The point lies above the line or right side of the line";
   else
      print "The point lies on the line"
End

示例代码

#include
#include
#include
#include
using namespace std;

const int L = 0;
const int H= 20;

int main(int argc, char **argv) {
   time_t seconds;
   time(&seconds);
   srand((unsigned int) seconds);

   int x1, x2, y1, y2;
   x1 = rand() % (H - L + 1) + L;
   x2 = rand() % (H - L + 1) + L;
   y1 = rand() % (H - L + 1) + L;
   y2 = rand() % (H - L + 1) + L;

   cout << "The Equation of the 1st line is : (" << (y2 - y1) << ")x+(" << (x1 - x2) << ")y+(" << (x2 * y1 - x1 * y2) << ") = 0\n";

   int x, y;
   cout << "\nEnter the point:";
   cin >>x;
   cin >>y;

   int s = (y2 - y1) * x + (x1 - x2) * y + (x2 * y1 - x1 * y2);
   if (s < 0)
      cout << "The point lies below the line or left side of the line";
   else if (s >0)
      cout << "The point lies above the line or right side of the line";
   else
      cout << "The point lies on the line";
      return 0;
}

输出结果

The Equation of the 1st line is : (7)x+(0)y+(-105) = 0

Enter the point:7
6
The point lies below the line or left side of the line

返回顶部
3162201930
czq8825@qq.com

应用上下测试来查找点相对于直线的位置的 C++ 程序

算法

示例代码

热门推荐

随机推荐