SGU106 (扩展欧几里得)

The equation

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

分析：给出一个方程ax+by=-c,当c%gcd(a,b)!=0的时候方程无解。
当c%gcd(a,b)==0时，判断一下a，b为0的情况，
当a和b都不为0时，设r=gcd(a,b),先用扩展欧几里得求出ax+by=r的一组特解(x0,y0),
那么原方程的一组特解是(x0*c/r,y0*c/r),
改写成ax0+by0=-c,
设n为整数，则ax0+n*(a*b/r)+by0-n*(a*b/r)=-c,
即a(x0+n*b/r)+b(y0-n*a/r)=-c,
那么x=x0+n*b/r，y=y0-n*a/r
由题目中x1<=x<=x2,y1<=y<=y2得，
x1<=x0+n*b/r<=x2,
y1<=y0-n*a/r<=y2,
解这不等式组就可以了。
扩展gcd的y=t-a/b*y;写成y=t-y*a/b，WA了好多次=_=......

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iostream>
using namespace std;
long long exgcd(long long a,long long b,long long &x,long long &y)
{
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    long long r=exgcd(b,a%b,x,y);
    long long t=x;
    x=y;
    y=t-a/b*y;//y=t-y*a/b是错的 
    return r;
}
int main()
{
    long long a,b,c,x1,x2,y1,y2;
    cin>>a>>b>>c>>x1>>x2>>y1>>y2;
    long long x,y;
    long long r=exgcd(a,b,x,y);
    c=-c;
    if(a==0||b==0)
    {
        if(a==0&&b==0)
        {
            if(c==0)
            printf("%lld
",(x2-x1+1)*(y2-y1+1));
            else printf("0
");
        }
        else if(a==0)
        {
            if(c%b==0&&c/b>=y1&&c/b<=y2)
            printf("%lld
",x2-x1+1);
            else printf("0
");
        }
        else
        {
            if(c%a==0&&c/a>=x1&&c/a<=x2)
            printf("%lld
",y2-y1+1);
            else printf("0
");
        }
        return 0;
    }
    if(c%r)    printf("0
"); 
    else
    {
        long long t=c/r;
        x*=t;y*=t;
        long long k=b/r;
        double L1,R1;
        if(k>0)
        {
            L1=(double)(x1-x)/(double)k;
            R1=(double)(x2-x)/(double)k;
        }
        else
        {
            L1=(double)(x2-x)/(double)k;
            R1=(double)(x1-x)/(double)k;
        }
        
        k=a/r;
        double L2,R2;
        if(k>0)
        {
            L2=(double)(y-y2)/(double)k;
            R2=(double)(y-y1)/(double)k;
        }
        else
        {
            L2=(double)(y-y1)/(double)k;
            R2=(double)(y-y2)/(double)k;
        }
        
        double L,R;
        L=max(L1,L2);
        R=min(R1,R2);
        
        long long ans=(long long)(floor(R+1e-6)-ceil(L-1e-6)+1);
        //printf("ans=%d
",ans);
        if(ans>0) printf("%lld
",ans);
        else printf("0
");
    }
    
    return 0;
}