SGU 106 The equation 扩展欧几里德

106. The equation

time limit per test: 0.25 sec.
memory limit per test: 4096 KB

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 -3
0 4
0 4

Sample Output

4
思路：ax+by=-c；
　　  扩展欧几里德求解；
　　  x=x0+b/gcd(a,b)*t;
　　　y=y0+a/gcd(a,b)*t;
     求x1<=x<=x2&&y1<=y<=y2的条件下，t的可行解；
　　　找到x的范围的t的可行解[lx,rx];
　　　同理                [ly,ry];
     ans=min(rx,ry)-max(lx,ly)+1;

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define esp 1e-13
const int N=1e3+10,M=1e6+1000,inf=1e9+10,mod=1000000007;
void extend_Euclid(ll a, ll b, ll &x, ll &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return;
    }
    extend_Euclid(b, a % b, x, y);
    ll tmp = x;
    x = y;
    y = tmp - (a / b) * y;
}
ll gcd(ll a,ll b)
{
    if(b==0)
        return a;
    return gcd(b,a%b);
}
int main()
{
    ll a,b,c;
    ll lx,rx;
    ll ly,ry;
    scanf("%I64d%I64d%I64d",&a,&b,&c);
    scanf("%I64d%I64d",&lx,&rx);
    scanf("%I64d%I64d",&ly,&ry);
    c=-c;
    if(lx>rx||ly>ry)
    {
        printf("0\n");
        return 0;
    }
    if (a == 0 && b == 0 && c == 0)
    {
        printf("%I64d\n",(rx-lx+1) * (ry-ly+1));
        return 0;
    }
    if (a == 0 && b == 0)
    {
        printf("0\n");
        return 0;
    }
    if (a == 0)
    {
        if (c % b != 0)
        {
            printf("0\n");
            return 0;
        }
        ll y = c / b;
        if (y >= ly && y <= ry)
        {
            printf("%I64d\n",rx - lx + 1);
            return 0;
        }
        else
        {
            printf("0\n");
            return 0;
        }
    }
    if (b == 0)
    {
        if (c % a != 0)
        {
            printf("0\n");
            return 0;
        }
        ll x = c / a;
        if (x >= lx && x <= rx)
        {
            printf("%I64d\n",ry - ly + 1);
            return 0;
        }
        else
        {
            printf("0\n");
            return 0;
        }
    }
    ll hh=gcd(abs(a),abs(b));
    if(c%hh!=0)
    {
        printf("0\n");
        return 0;
    }
    else
    {
        ll x,y;
        extend_Euclid(abs(a),abs(b),x,y);
        x*=(c/hh);
        y*=(c/hh);
        if(a<0)
            x=-x;
        if(b<0)
            y=-y;
        a/=hh;
        b/=hh;
        ll tlx,trx,tly,trry;
        if(b>0)
        {
            ll l=lx-x;
            tlx=l/b;
            if(l>=0&&l%b)
                tlx++;
            ll r=rx-x;
            trx=r/b;
            if(r<0&&r%b)
                trx--;
        }
        else
        {
            b=-b;
            ll l=x-rx;
            tlx=l/b;
            if(l>=0&&l%b)
                tlx++;
            ll r=x-lx;
            trx=r/b;
            if(r<0&&r%b)
                trx--;
        }
        if(a>0)
        {
            ll l=-ry+y;
            tly=l/a;
            if(l>=0&&l%a)
                tly++;
            ll r=-ly+y;
            trry=r/a;
            if(r<0&&r%a)
                trry--;
        }
        else
        {
            a=-a;
            ll l=ly-y;
            tly=l/a;
            if(l>=0&&l%a)
                tly++;
            ll r=ry-y;
            trry=r/a;
            if(r<0&&r%a)
                trry--;
        }
        printf("%I64d\n",(max(0LL,min(trry,trx)-max(tly,tlx)+1)));
        return 0;
    }
    return 0;
}