The equation （扩展欧几里得）题解

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：

又是一道很明显的exgcd题，这次做完，感觉对exgcd了解更加多了。

对a要进行符号判断，负号就要变为正号，相对应的区间x1，x2也要取对称区间；b同理。c变换a，b也要一起变换（二元一次方程）。其他的可以参考之前写过的题：循环狂魔 和 青蛙也要找女朋友

一道解二元一次方程的题，最后一点并集那里画个图应该就能解决了。中间还有一些判断要分布讨论。

又找到一个ceil（）用来求向上取整（里面必须要double，和floor（）一样，不然提交就会CE...）

代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<queue>
#include<cmath>
#include<string>
#include<map>
#include<stack> 
#include<set>
#include<vector>
#include<iostream>
#include<algorithm>
#include<sstream>
#define INF 0x3f3f3f3f
#define ll long long
const int N=10005;
const ll MOD=998244353;
using namespace std;
ll ex_gcd(ll a,ll b,ll &x,ll &y){
	ll d,t;
	if(b==0){
		x=1;
		y=0;
		return a;
	}
	d=ex_gcd(b,a%b,x,y);
	t=x-a/b*y;
	x=y;
	y=t;
	return d;
}
int main(){
	ll a,b,c,x1,x2,y1,y2,x,y;
	cin>>a>>b>>c>>x1>>x2>>y1>>y2;
	c=-c;
	if(c<0){
		c=-c;
		a=-a;
		b=-b;
	}
	if(a<0){
		a=-a;
		swap(x1,x2);
		x1=-x1;
		x2=-x2; 
	}
	if(b<0){
		b=-b;
		swap(y1,y2);
		y1=-y1;
		y2=-y2; 
	}
	ll d=ex_gcd(a,b,x,y);
	if(a==0 || b==0){	//ax+by=-c 
		if(a==0 && b==0){
			if(c==0){
				cout<<(x2-x1+1)*(y2-y1+1)<<endl;
				return 0; 
			}
			else{
				cout<<0<<endl;
				return 0;
			}
		}
		else if(a==0){
			if(c%b==0 && c/b>=y1 && c/b<=y2){
				cout<<(x2-x1+1)<<endl;
				return 0;
			}
			else{
				cout<<0<<endl;
				return 0;
			}
		}
		else if(b==0){
			if(c%a==0 && c/a>=x1 && c/a<=x2){
				cout<<(y2-y1+1)<<endl;
				return 0;
			}
			else{
				cout<<0<<endl;
				return 0;
			}
		}
	}
	x=x*c/d;
	y=y*c/d;
	ll k1=b/d,k2=a/d;
	if(c%d!=0){
		cout<<0<<endl;
		return 0;
	}
	else{
		ll r=min(floor((x2-x)*1.0/k1),floor((y-y1)*1.0/k2)) ,l=max(ceil((x1-x)*1.0/k1),ceil((y-y2)*1.0/k2));
		if(r>=l){
			cout<<r-l+1<<endl;
		} 
		else{
			cout<<0<<endl;
		}
	}
	return 0;
}