The equation(exgcd+解不等式方程组)

The equation

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There is an equation ax + by + c = 0. Given a,b,c,x1,x2,y1,y2 you must determine, how many integer roots of this equation are satisfy to the following conditions : x1<=x<=x2,   y1<=y<=y2. Integer root of this equation is a pair of integer numbers (x,y).

Input

Input contains integer numbers a,b,c,x1,x2,y1,y2 delimited by spaces and line breaks. All numbers are not greater than 10⁸ by absolute value.

Output

Write answer to the output.

Sample Input

1 1 -30 40 4

Sample Output

4

 
#include <bits/stdc++.h>using namespace std;typedef long long ll;ll exgcd(ll a,ll b,ll &x,ll &y){    if (b==0)    {        x=1;        y=0;        return a;    }    ll d =exgcd(b,a%b,x,y);    ll t=x ;    x=y;    y=t-a/b*y;    return d;}//exgcd 板子题，不过后台数据还是可以的，/*****特殊情况的三种判断，a==0,b==0;a==0,b!=0,a==0,b!=0;第四种情况，exgcd,本身自己大算在第四中情况中先找到x的范围，在对应的跑完对应y的范围，结果第十二组测试数据卡超时只好找x的范围，y的范围，求其交集，即为最后的答案。。自己wa进30发。。。。：（最后想说的是，学会了两个函数，floor（），向下取整，ceil()向上取整，举例[2.5,4.5]对应的范围应该是[3,4],[-3.5,5.5]对应的范围应该是[-3,5];所以上去下界，下去上届。****/int main (){    ll a,b,c,l_x,r_x,xx,xxx,x,y,yy,yyy;    cin>>a>>b>>c>>xx>>xxx>>yy>>yyy;    ll ans=0;    c=-c;    if (a==0&&b==0)    {         if (c==0)//最 zz的一次，竟然不加这个判断，不加的话，等式不成立。。。。。        ans =(xxx-xx+1)*(yyy-yy+1);    }    else if (a==0&&b!=0)    {        if (c%b==0 && (c/b)>=yy && (c/b)<=yyy)            ans=(xxx-xx+1);    }    else if (a!=0&&b==0)    {        if (c%a==0&&(c/a)>=xx&&(c/a)<=xxx)            ans=(yyy-yy+1);    }    else    {        ll d=exgcd(a,b,x,y);        double  aa,bb;        if (c%d==0)        {            x=c/d*x;            y=c/d*y;            aa=-(a/d);            bb=b/d;            if(bb>0)            {                l_x=ceil(((double)xx-x)/bb);                r_x=floor(((double)xxx-x)/bb);            }            else            {                l_x=ceil(((double)xxx-x)/bb);                r_x=floor(((double)xx-x)/bb);            }            ll l_y,r_y;            if(aa>0)            {                l_y=ceil(((double)yy-y)/aa);                r_y=floor(((double)yyy-y)/aa);            }            else            {                l_y=ceil(((double)yyy-y)/aa);                r_y=floor(((double)yy-y)/aa);            }             ans =min(r_x,r_y)-max(l_x,l_y)+1;             if(ans<0)                    ans=0;        }    }        cout<<ans<<endl;    return 0;}