[codevs1213]解的个数（数论，扩欧）

已知整数x,y满足如下面的条件:

ax+by+c = 0

p<=x<=q

r<=y<=s

求满足这些条件的x,y的个数。

#include<iostream>#define ll long longusing namespace std;ll n,a,b,c,p,q,r,s; ll exgcd(ll a,ll b,ll &x,ll &y){    if (b==0){        x=1;        y=0;        return a;    }    ll r=exgcd(b,a%b,x,y);    ll t=x;    x=y;    y=t-a/b*y;    return r;}int pd(ll x,ll l,ll r){    if (x>=l&&x<=r) return 1;    return 0;}int main(){    cin>>n;    for (int i=1;i<=n;i++){        ll ans=0;        cin>>a>>b>>c>>p>>q>>r>>s;        c=-c;        ll x=0,y=0;        if (a==0&&b==0&&c!=0) {            cout<<"0"<<"\n";            continue;        }        if (p>q||r>s){            cout<<"0"<<"\n";            continue;        }        if (a==0&&b==0&&c==0){            cout<<(q-p+1)*(s-r+1)<<"\n";            continue;        }        int d=exgcd(a,b,x,y);        if (c%d!=0){            cout<<"0\n";            continue;        }        x=x*c/d;        y=y*c/d;        int t1,t2;        if (b==0){            if (x>=p&&x<=q) cout<<"1\n";            else cout<<"0\n";            continue;        }        else if ((float)d/b>0){            t1=(p-x)*d/b;            t2=(q-x)*d/b;        }        else{            t1=(q-x)*d/b;            t2=(p-x)*d/b;        }        for (ll k=t1;k<=t2;k++)            if (pd(x+k*b/d,p,q)&&pd(y-k*a/d,r,s)) ans++;        cout<<ans<<endl;    }}