sgu106 求直线在某一矩形范围内经过的整点，扩展gcd

sgu106  求直线在某一矩形范围内经过的整点，扩展gcd

X - The equation

Time Limit:250MS     Memory Limit:4096KB     64bit IO Format:%I64d & %I64u

Submit Status Practice SGU 106

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 -3
0 4
0 4

Sample Output

4
题意：求解ax+bx+c=0在[x1,x2][y1,y2]经过的整点个数。
思路：扩展gcd解出一个特解，再根据[x1,x2][y1,y2]解出两个k的范围，取交集即可。一个细节，先取交集再取整WA，先取整再取交集就过了，不知为何。。一个技巧，对取整数闭区间可以对两个double边界，下界向上取整ceil，上界向下取整floor，即可取到整数边界的闭区间。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<string>
#include<math.h>
#include<cctype>
#define ll long long
#define REP(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define REPP(i,a,b,t) for(int (i)=(a);(i)<=(b);(i)+=(t))
#define PII pair<int,int>
#define fst first
#define snd second
#define MP make_pair
#define PB push_back
#define RI(x) scanf("%d",&(x))
#define RII(x,y) scanf("%d%d",&(x),&(y))
#define RIII(x,y,z) scanf("%d%d%d",&(x),&(y),&(z))
#define DRI(x) int (x);scanf("%d",&(x))
#define DRII(x,y) int (x),(y);scanf("%d%d",&(x),&(y))
#define DRIII(x,y,z) int (x),(y),(z);scanf("%d%d",&(x),&(y),&(z))
#define RS(x) scanf("%s",s)
#define RSS(x,y) scanf("%s%s",x,y)
#define DRS(x) char x[maxn];scanf("%s",x)
#define DRSS(x,y) char x[maxn],y[maxn];scanf("%s%s",x,y)
#define MS0(a) memset((a),0,sizeof((a)))
#define MS1(a) memset((a),-1,sizeof((a)))
#define MS(a,b) memset((a),(b),sizeof((a)))
#define ALL(v) v.begin(),v.end()
#define SZ(v) (v).size()

using namespace std;

const int maxn=1000100;
const int INF=(1<<29);
const double EPS=0.0000000001;
const double Pi=acos(-1.0);

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=y;
    y=x-a/b*y;
    x=t;
    return r;
}

int main()
{
    ll a,b,c,x1,x2,y1,y2,x,y;
    while(cin>>a>>b>>c>>x1>>x2>>y1>>y2){
        c=-c;
        if(a==0&&b==0)
            if(c==0) cout<<(x2-x1+1)*(y2-y1+1)<<endl;
            else cout<<0<<endl;
        else if(a==0)
            if(c%b==0&&y1<=c/b&&c/b<=y2) cout<<x2-x1+1<<endl;
            else cout<<0<<endl;
        else if(b==0)
            if(c%a==0&&x1<=c/a&&c/a<=x2) cout<<y2-y1+1<<endl;
            else cout<<0<<endl;
        else{
            ll d=exgcd(a,b,x,y);
            if(c%d) cout<<0<<endl;
            else{
                x*=c/d;y*=c/d;
                double L1=(x1-x)*1.0*d/b,R1=(x2-x)*1.0*d/b;
                if(L1>R1) swap(L1,R1);
                double L2=(y-y1)*1.0*d/a,R2=(y-y2)*1.0*d/a;
                if(L2>R2) swap(L2,R2);
                ll Ld1=ceil(L1),Rd1=floor(R1);
                ll Ld2=ceil(L2),Rd2=floor(R2);
                ll L=max(Ld1,Ld2),R=min(Rd1,Rd2);
                if(L>R) cout<<0<<endl;
                else cout<<R-L+1<<endl;
            }
        }
    }
    return 0;
}