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Acaini.yaoyaozii — Sun, 04 Oct 2009 10:38:00 GMT

一.叉积

   �?a(x1,y1), b(x2,y2)
二维�Q�a x b=x1*y2-x2*y1

   设p0(x0,y0), p1(x1,y1) ,p2(x2,y2), p3(x3,y3)
  < p0p1>= (p1-p0) =(x1-x0,y1-y0);
   = (p3-p2) =(x3-x2,y3-y2);
    x =(p1-p0) x (p3-p2) = (x1-x0)*(y3-y2)- (x3-x2)* (y1-y0);

(p1-p0) x (p3-p2)�l�果的意�?/span>:

�?/span>: �?/span>��时�?/span>(0,pi)�?/span>

�?/span>: �?/span>逆时�?/span>(0,pi)�?/span>

0 : , ��q��,夹角�?/span>0�?/span>pi

double xmult(point p0,point p1,point p2){
return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}

可用于判p2�?lt;p0 p1>的关�p?br>

double xmult(point p0,point p1,point p2,point p3){
return (p1.x-p0.x)*(p3.y-p2.y)-(p3.x-p2.x)*(p1.y-p0.y);
}

可用于判, 位置关系

double xmult(line u,line v){
return (u.b.x-u.a.x)*(v.b.y-v.a.y)-(v.b.x-v.a.x)*(u.b.y-u.a.y);
}

//计算两直�U�交�?注意事先判断直线是否�q��!
point intersection(line u,line v){
    point ret=u.a;
    double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))
            /((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));
    ret.x+=(u.b.x-u.a.x)*t;
    ret.y+=(u.b.y-u.a.y)*t;
    return ret;
}

int relat_and_intersection(line u,line v,point &ret)
{
    /*
    LINE 重合
    NONE �q��
    INTERSECT �怺��q�返回交�?br>    */
    double a1,b1,c1,a2,b2,c2;
    //�p�L��
    a1=u.a.y-u.b.y;
    b1=-(u.a.x-u.b.x);
    c1=u.a.x*u.b.y-u.a.y*u.b.x;

    a2=v.a.y-v.b.y;
    b2=-(v.a.x-v.b.x);
    c2=v.a.x*v.b.y-v.a.y*v.b.x;

    if(fabs(a1*b2-a2*b1)<eps)
    {
        if( fabs(a1*c2-a2*c1)<eps && fabs(b1*c2-b2*c1)<eps )
            return LINE;
        else return NONE;
    }
    else
    {
        point ret;
        ret.x=(b1*c2-b2*c1)/(a1*b2-a2*b1);
        ret.y=(a2*c1-a1*c2)/(a1*b2-a2*b1);
        return INTERSECT;
    }
}

Acaini.yaoyaozii 2009-10-04 18:38 发表评论

KMP

Acaini.yaoyaozii — Thu, 30 Jul 2009 09:22:00 GMT

#include<iostream>
using namespace std;
#define M 1000
//int kmp(char *t,char *p,int pos)
int kmp(char *t,char *p)
{
    //p模式�Ԍ��t��M��
    //预处�?/span>
    int next[M];
    //memset(next,0,sizeof(next));
    int  i,j,
        lent=strlen(t),
        lenp=strlen(p);
    next[0]=-1;
    i=0;j=-1;
    while(i<lenp-1)
    {
        if(j==-1 || p[i]==p[j])
        {
            ++i;++j;
            if(p[i]!=p[j]) next[i]=j;
            else next[i]=next[j];
            //next[i]=j;
        }
        else j=next[j];
    }
    //匚w��
    i=0;j=0;
    while(i<lent && j<lenp)
    {
        if(j==-1 || t[i]==p[j]) {++i;++j;}
        else j=next[j];
    }
    if(j==lenp) return i-lenp;
    else return -1;
}

int main()
{
    char t[100],p[100];
    while(cin>>t>>p)
        cout<<kmp(t,p)<<endl;
    return 0;
}

Acaini.yaoyaozii 2009-07-30 17:22 发表评论

Acaini.yaoyaozii — Sun, 26 Jul 2009 02:39:00 GMT

#define M 10000
bool prime[M];
int pri[M];
void prime()
{
    //1表示不是素数�Q?表示是素�?br>    //memset(prime,0,sizeof(prime));
    int i,j,
        k=0;
    prime[0]=prime[1]=1;
    for(i=2;i<M;i++)
        if(prime[i]==0)
        {
            //pri[k++]=i;
            for(j=2*i;j<M;j+=i)
                prime[j]=1;
        }
}

Acaini.yaoyaozii 2009-07-26 10:39 发表评论