You can Solve a Geometry Problem too

Many geometry（几何）problems were designed in the ACM/ICPC. And now, I also prepare a geometry problem for this final exam. According to the experience of many ACMers, geometry problems are always much trouble, but this problem is very easy, after all we are now attending an exam, not a contest :) Give you N (1<=N<=100) segments（线段）, please output the number of all intersections（交点）. You should count repeatedly if M (M>2) segments intersect at the same point. Note: You can assume that two segments would not intersect at more than one point.

Input contains multiple test cases. Each test case contains a integer N (1=N<=100) in a line first, and then N lines follow. Each line describes one segment with four float values x1, y1, x2, y2 which are coordinates of the segment’s ending. A test case starting with 0 terminates the input and this test case is not to be processed.

For each case, print the number of intersections, and one line one case.

20.00 0.00 1.00 1.000.00 1.00 1.00 0.0030.00 0.00 1.00 1.000.00 1.00 1.00 0.0000.00 0.00 1.00 0.000

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#include 
   
    //author:XXYYstruct Point{	double x1,y1;	double x2,y2;}p[110];int jug(Point a,Point b){	double x=b.y1-(b.x1*(a.y1-a.y2)+(a.x1*a.y2-a.x2*a.y1))/(a.x1-a.x2);	double y=b.y2-(b.x2*(a.y1-a.y2)+(a.x1*a.y2-a.x2*a.y1))/(a.x1-a.x2);	if(x*y>0)		return 0;//题意说明该题两线段相交构成十字，不会出现T字	return 1;}int main(){	int n,i,j,num;	while(scanf("%d",&n),n){		for(i=0;i