SGU 106 The equation（扩欧）

Description

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

思路：

1.如果x1>x2或者1>y2，不符合题意，无解.

2.只要a,b不同时为0，gcd(a,b)!=0,就可以带入扩欧函数,求得一组解，所以先判断a=0,b=0的情况，然后把a,b,带入函数，求得x,y,再讨论a,b分别为0的情况.

4.根据xx=x+k*d,yy=y-k*a(k属于Z)，带入x,y的上下界，求得整数k1,k2,k3,k4,使得k1<k2,k3<k4,然后求[k1,k2],[k3,k4]的交集.

注意：

ceil()为向上取整，floor()为向下取整.

代码：

</pre><pre name="code" class="cpp">#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>using namespace std;typedef long long ll;ll mmax(ll a,ll b){return a>b?a:b;}ll mmin(ll a,ll b){return a<b?a:b;}ll exgcd(ll a,ll b,ll &x,ll &y){    if(!b) {x=1;y=0;return a;}    ll d=exgcd(b,a%b,y,x);    y-=a/b*x;    return d;}int main(){    ll a,b,c,d,x1,x2,y1,y2,x,y,k1,k2,k3,k4;    cin>>a>>b>>c>>x1>>x2>>y1>>y2;c=-c;    if(x1>x2||y1>y2)    {        cout<<0<<endl;        return 0;    }    if(a==0&&b==0)    {        if(c==0)            cout<<(x2-x1+1)*(y2-y1+1)<<endl;        else            cout<<0<<endl;        return 0;    }    d=exgcd(a,b,x,y);    if(c%d)    {        cout<<0<<endl;        return 0;    }    a/=d,b/=d,c/=d;    x*=c,y*=c;    if(a==0)    {        if(y>=y1&&y<=y2)            cout<<x2-x1+1<<endl;        else            cout<<0<<endl;        return 0;    }    if(b==0)    {        if(x>=x1&&x<=x2)            cout<<y2-y1+1<<endl;        else            cout<<0<<endl;        return 0;    }    double aa=-a,bb=b,xx=x,yy=y,root_x1,root_x2,root_y1,root_y2;    root_x1=(x1-xx)/bb,root_x2=(x2-xx)/bb,root_y1=(y1-yy)/aa,root_y2=(y2-yy)/aa;    if(root_x1>root_x2) swap(root_x1,root_x2);    if(root_y1>root_y2) swap(root_y1,root_y2);    k1=ceil(root_x1),k2=floor(root_x2),k3=ceil(root_y1),k4=floor(root_y2);    cout<<mmin(k2,k4)-mmax(k1,k3)+1<<endl;    return 0;}