几何公式

两条直线是否相交

//叉积
double mult(Point a, Point b, Point c)
{
    return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
}

//aa, bb为一条线段两端点 cc, dd为另一条线段的两端点 相交返回true, 不相交返回false
bool intersect(Point aa, Point bb, Point cc, Point dd)
{
    if ( max(aa.x, bb.x)<min(cc.x, dd.x) )
    {
        return false;
    }
    if ( max(aa.y, bb.y)<min(cc.y, dd.y) )
    {
        return false;
    }
    if ( max(cc.x, dd.x)<min(aa.x, bb.x) )
    {
        return false;
    }
    if ( max(cc.y, dd.y)<min(aa.y, bb.y) )
    {
        return false;
    }
    if ( mult(cc, bb, aa)*mult(bb, dd, aa)<0 )
    {
        return false;
    }
    if ( mult(aa, dd, cc)*mult(dd, bb, cc)<0 )
    {
        return false;
    }
    return true;
}

圆周率计算代码

#include<stdio.h>
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF&&n!=0)
    {
        int const N=7200;
        int const M=10000;
        int const  B=10000;
        int const L=4;
        int s[M/L];
        int r1[N]= {0},r2[N]= {0},d1[N]= {0},d2;
        int r3[N]= {0},r4[N]= {0},d3[N]= {0},d4;
        int i,k,t,p=0,mp=M/L/20;
        r1[0]=1;
        r1[1]=3;
        r3[0]=4;
        printf("正在计算,请等待
____________________
");
        for(k=0; k<M/L; ++k)
        {
            t=r1[0]*B;
            d1[0]=t/0x5;
            r1[0]=t%0x5;
//
            t=r3[0]*B;
            d3[0]=t/0xEF;
            r3[0]=t%0xEF;
            s[k]=d1[0]-d3[0];
            int tag=0;
            for(i=1; i<N; ++i)
            {
                t=r1[i]*B+d1[i-1];
                d1[i]=t/0x19;
                r1[i]=t%0x19;
                t=r2[i]*B+d1[i];
                d2=t/(2*i+1);
                r2[i]=t%(2*i+1);
//
                t=r3[i]*B+d3[i-1];
                d3[i]=t/0xDF21;
                r3[i]=t%0xDF21;
                t=r4[i]*B+d3[i];
                d4=t/(2*i+1);
                r4[i]=t%(2*i+1);
                if(tag)
                {
                    s[k]+=(d2-d4);
                    tag=0;
                }
                else
                {
                    s[k]+=(d4-d2);
                    tag=1;
                }
            }
            if(p==mp)
            {
                printf(">");
                p=0;
            }
            else
                p++;
        }
        for(i=M/L-1; i>=0; i--)
        {
            while(s[i]>=B)
            {
                s[i-1]++;
                s[i]-=B;
            }
            while(s[i]<0)
            {
                s[i-1]--;
                s[i]+=B;
            }
        }
        printf("
pi=3.
");
        for(i=0; i<M/L; ++i)
            printf("%d",s[i]);
        printf("
");
    }
    return 0;

}

求圆心

void count(double x1,double y1,double x2,double y2,double x3,double y3)
{
    double a,b,c,d,e,f;
    a=2*(x2-x1);
    b=2*(y2-y1);
    c=x2*x2+y2*y2-x1*x1-y1*y1;
    d=2*(x3-x2);
    e=2*(y3-y2);
    f=x3*x3+y3*y3-x2*x2-y2*y2;
    x=(b*f-e*c)/(b*d-e*a);
    y=(d*c-a*f)/(b*d-e*a);
    r=sqrt((x-x1)*(x-x1)+(y-y1)*(y-y1));

}