SGU106[The equation]

数学题。扩展欧几里得解模同余方程。


CODE:


/*


AUTHOR: Su Jiao


DATE: 2010-8-26


DESCRIPTION:


$DESCRIPTION


*/


#include <iostream>


#include <math.h>


using namespace std;


 


#define lli long long int


#define max(a,b) ((a)>(b)?(a):(b))


#define min(a,b) ((a)<(b)?(a):(b))


#define ceil_div(a,b) ceil(double(a)/double(b))


#define floor_div(a,b) floor(double(a)/double(b))


 


lli gcd(lli a,lli b,lli& x,lli& y)


{


    //ax+by=gcd(a,b)


    if (b==0)


    {


       x=1,y=0;


       return a;


    }


    else


    {


        lli g=gcd(b,a%b,y,x);


        //b*y+a%b*x=gcd(b,a%b)=gcd(a,b)


        // b*y+a%b*x


        //=b*y+(a-a/b*b)*x


        //=b*y-a/b*b*x+a*x


        //=a*x+b*(y-a/b*x)


        y-=a/b*x;


        return g;


    }


}


lli get_answer(lli a,lli b,lli c,lli x1,lli x2,lli y1,lli y2)


{


    //ax+by+c=0


    if (!(x1<=x2&&y1<=y2)) return 0;


    else if (a==0&&b==0)


         return (c==0)*(x2-x1+1)*(y2-y1+1);


    else if (a==0)


    {


         lli y=-c/b;


         return ((b*y+c==0)&&(y1<=y&&y<=y2))*(x2-x1+1);


    }


    else if (b==0)


    {


         lli x=-c/a;


         return ((a*x+c==0)&&(x1<=x&&x<=x2))*(y2-y1+1);


    }


    else


    {


        //a*x = c (mod b)


        lli x,y,g;


        g=gcd(a,b,x,y);//ax+by=g


        if (c%g==0)


        {


           a/=g,b/=g,c/=g;//ax+by=1


           x*=-c,y*=-c;//ax*(-c)+by*(-c)=-c          


           //set of root = {(x+b*k,y-a*k)}          


           lli rxa,rxb;


           if (b>0) rxa=ceil_div(x1-x,b),rxb=floor_div(x2-x,b);


           else rxa=ceil_div(x2-x,b),rxb=floor_div(x1-x,b);


           lli rya,ryb;


           if (a<0) rya=ceil_div(y-y1,a),ryb=floor_div(y-y2,a);


           else rya=ceil_div(y-y2,a),ryb=floor_div(y-y1,a);


           lli ra,rb;


           ra=max(rxa,rya);


           rb=min(rxb,ryb);


           if (ra<=rb) return (rb-ra+1);


           else return 0;


        }


        else return 0;


    }


}


 


int main()


{


    lli a,b,c,x1,x2,y1,y2;


    cin>>a>>b>>c>>x1>>x2>>y1>>y2;


    cout<<get_answer(a,b,c,x1,x2,y1,y2)<<endl;


}