you can Solve a Geometry Problem too

Problem Description

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

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.

Output

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

Sample Input

2

0.00 0.00 1.00 1.00

0.00 1.00 1.00 0.00

3

0.00 0.00 1.00 1.00

0.00 1.00 1.00 0.00

0.00 0.00 1.00 0.00

0

/*判断AB和CD两线段是否有交点：
同时满足两个条件：(‘x‘表示叉积)
　　　　1．C点D点分别在AB的两侧．(向量(ABxAC)*(ABxAD)<=0)
             2．A点和B点分别在CD两侧．(向量(CDxCA)*(CDxCB)<=0)*/
/*数据稍微多点我就写错了，不求快但求稳*/
#include<stdio.h>
#include<string.h>
int fun(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4)
{
    double d1=(x2-x1)*(y4-y1)-(y2-y1)*(x4-x1),d2=(x2-x1)*(y3-y1)-(y2-y1)*(x3-x1);
    if(d2*d1<=0)
    {
        double d3=(x4-x3)*(y1-y3)-(y4-y3)*(x1-x3),d4=(x4-x3)*(y2-y3)-(x2-x3)*(y4-y3);
        if(d3*d4<=0)
            return 1;
        return 0;
    }
    else
        return 0;
}

int main()
{
    double a[101][4];
    int n,i,j;
    while(~scanf("%d",&n)&&n)
    {
        memset(a,0,sizeof(a));
        int count=0;
        for(i=1;i<=n;i++)
            scanf("%lf%lf%lf%lf",&a[i][0],&a[i][1],&a[i][2],&a[i][3]);
        for(j=1;j<=n;j++)
            for(i=j+1;i<=n;i++)
                count+=fun(a[j][0],a[j][1],a[j][2],a[j][3],a[i][0],a[i][1],a[i][2],a[i][3]);
    printf("%d\n",count);
    }
}
                    

Sample Output

1 3