在具有给定两端的线段上找到整数点

给定XY空间中的两个点pointU和pointV ，我们需要找到一个具有整数坐标的点，该点位于通过点pointU和pointV的直线上。
例子：

If  pointU = (1, -1 and pointV = (-4, 1)
then equation of line which goes 
through these two points is,
2X + 5Y = -3
One point with integer co-ordinate which
satisfies above equation is (6, -3)

我们可以看到，一旦找到线方程，就可以将此问题视为扩展Euclid算法问题，其中我们知道AX + BY = C中的A，B，C，并希望从中找出X和Y的值。等式。
在上面的扩展Euclid方程中，C是A和B的gcd，因此从给定的两点找出线方程后，如果C不是gcd(A，B)的倍数，则可以得出结论：指定的行。如果C是g的倍数，那么我们可以按比例增加已建立的X和Y系数以满足实际方程式，这将是我们的最终答案。

//   C++ program to get Integer point on a line
#include 
using namespace std;
  
//  Utility method for extended Euclidean Algorithm
int gcdExtended(int a, int b, int *x, int *y)
{
    // Base Case
    if (a == 0)
    {
        *x = 0;
        *y = 1;
        return b;
    }
  
    int x1, y1; // To store results of recursive call
    int gcd = gcdExtended(b%a, a, &x1, &y1);
  
    // Update x and y using results of recursive
    // call
    *x = y1 - (b/a) * x1;
    *y = x1;
  
    return gcd;
}
  
//  method prints integer point on a line with two
// points U and V.
void printIntegerPoint(int c[], int pointV[])
{
    //  Getting coefficient of line
    int A = (pointU[1] - pointV[1]);
    int B = (pointV[0] - pointU[0]);
    int C = (pointU[0] * (pointU[1] - pointV[1]) +
             pointU[1] * (pointV[0] - pointU[0]));
  
    int x, y;  // To be assigned a value by gcdExtended()
    int g = gcdExtended(A, B, &x, &y);
  
    // if C is not divisible by g, then no solution
    // is available
    if (C % g != 0)
        cout << "No possible integer point\n";
  
    else
  
        //  scaling up x and y to satisfy actual answer
        cout << "Integer Point : " << (x * C/g) << " "
             << (y * C/g) << endl;
}
  
//   Driver code to test above methods
int main()
{
    int pointU[] = {1, -1};
    int pointV[] = {-4, 1};
  
    printIntegerPoint(pointU, pointV);
    return 0;
}

输出：

Integer Point : 6 -3