SGU 106 The euqation

坑點：在處理閉區間邊界時，涉及到取ceil和floor的問題，當a>0，b>0時，a/b取floor,可以用(a-1)/b+1取ceil,而a<0，b>0時，a/b取ceil，可以用(a+1)/b-1取floor，a=0特殊考慮，直接a/b就行。 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
void ExEuclid(LL a,LL b,LL &x,LL &y,LL &d)
{
    if(b==0)
    {
        d=a,x=1,y=0;
        return ;
    }
    else
    {
        ExEuclid(b,a%b,x,y,d);
        LL t=x;
        x=y;
        y=t-a/b*y;
        return ;
    }
}
LL Work(LL a,LL b,LL c,LL x1,LL x2,LL y1,LL y2)
{
    LL l1=x1,r1=x2,l2=y1,r2=y2;
    if(a==0&&b==0)
    {
        if(c!=0)
            return 0;
        return (x2-x1+1)*(y2-y1+1);
    }
    if(a==0)
    {
        if(c%b!=0)
            return 0;
        if((-c)/b<=y2&&(-c)/b>=y1)
            return (x2-x1+1);
        return 0;
    }
    if(b==0)
    {
        if(c%a!=0)
            return 0;
        if((-c)/a<=x2&&(-c)/a>=x1)
            return (y2-y1+1);
        return 0;
    }
    if(a<0)
    {
        a=-a;
        swap(l1,r1);
        l1=-l1,r1=-r1;
    }
    if(b<0)
    {
        swap(l2,r2);
        l2=-l2,r2=-r2;
        b=-b;
    }
    LL x0,y0,d;
    ExEuclid(a,b,x0,y0,d);
    if(c%d!=0)
        return 0;
    c=(-c)/d,a/=d,b/=d;
    l1-=c*x0,r1-=c*x0;
    if(l1<=0)     //注意這裏邊界的處理，l1/=b使得絕對值減小，當l1爲正數時起到floor作用，當l1爲負數時起到ceil作用
        l1/=b;
    else
        l1=(l1-1)/b+1;    //只有當l1>=1時才能取到ceil作用
    if(r1<0)
    {
        r1=(r1+1)/b-1;
    }
    else
        r1/=b;
    l2-=c*y0,r2-=c*y0;
    swap(l2,r2);
    l2=-l2,r2=-r2;
    if(l2<=0)
        l2/=a;
    else
        l2=(l2-1)/a+1;
    if(r2<0)
    {
        r2=(r2+1)/a-1;
    }
    else
        r2/=a;
    LL l=l1,r=r1;
    if(l2>l1)
        l=l2;
    if(r2<r1)
        r=r2;
    if(l<=r)
        return r-l+1;
    else
        return 0;
}
int main()
{
    LL a,b,c,x1,y1,x2,y2;
    while(scanf("%lld %lld %lld %lld %lld %lld %lld",&a,&b,&c,&x1,&x2,&y1,&y2)!=EOF)
    {
        if(x1>x2||y1>y2)
            printf("0\n");
        else
            printf("%lld\n",Work(a,b,c,x1,x2,y1,y2));
    }
    return 0;
}