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+/*
+Challenge 101
+
+TASK #2 › Origin-containing Triangle
+Submitted by: Stuart Little
+You are given three points in the plane, as a list of six co-ordinates:
+A=(x1,y1), B=(x2,y2) and C=(x3,y3).
+
+Write a script to find out if the triangle formed by the given three
+co-ordinates contain origin (0,0).
+
+Print 1 if found otherwise 0.
+*/
+
+#include <iostream>
+
+typedef struct Point {
+    double x, y;
+} Point;
+
+double sign(Point p1, Point p2, Point p3) {
+    return (p1.x - p3.x) * (p2.y - p3.y)
+        - (p2.x - p3.x) * (p1.y - p3.y);
+}
+
+int point_in_triangle(Point p0, Point p1, Point p2, Point p3) {
+    double d1 = sign(p0, p1, p2);
+    double d2 = sign(p0, p2, p3);
+    double d3 = sign(p0, p3, p1);
+
+    bool has_neg = (d1 < 0.0) || (d2 < 0.0) || (d3 < 0.0);
+    bool has_pos = (d1 > 0.0) || (d2 > 0.0) || (d3 > 0.0);
+
+    if (!(has_neg && has_pos))
+        return 1;
+    else
+        return 0;
+}
+
+int main(int argc, char* argv[]) {
+    if (argc != 7) {
+        std::cerr << "Usage: ch-2 x1 y1 x2 y2 x3 y3" << std::endl;
+        exit(EXIT_FAILURE);
+    }
+    else {
+        Point p[3], p0;
+        for (int i = 0; i < 3; i++) {
+            p[i].x = atof(argv[i * 2 + 1]);
+            p[i].y = atof(argv[i * 2 + 2]);
+        }
+        p0.x = p0.y = 0.0;
+        std::cout << point_in_triangle(p0, p[0], p[1], p[2]) << std::endl;
+    }
+}