you can Solve a Geometry Problem too

/*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);
    }
}