应用上下测试来查找点相对于直线的位置的 C++ 程序
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输出结果#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; }
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