[codevs1213]解的个数 二分 + exgcd

题目←

扩欧，求不定方程解的个数
我们已经知道通解x0 = x +- (b/gcd(a,b))*t，那只要知道有多少个t使x在题目给定的范围中就行了
但还有y
怎么办？求交集！
分别二分确定在x取值范围内合法的t的范围和在y取值范围内合法的t的范围
然后交一下
值得一提的是，对于同一个t，由exgcd求出的x，y而找到的一组通解为
x + (b/gcd(a,b)) * t，y - (a/gcd(a,b)) * t
或
x - (b/gcd(a,b)) * t，y + (a/gcd(a,b)) * t

当a，b同号时，应注意t是互为相反数的
对此把某一组解的范围 *= -1就好了

还有一种解法是划定范围+枚举，时间上不如二分但代码更简洁

注意特判最后几组a,b == 0的数据……

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
LL A,B,C,P,Q,R,S;
int n;
LL a,b,c,m,x,y,G;
LL Gcd(LL a,LL b){
	return b ? Gcd(b,a%b) : a;
}
void exgcd(LL a,LL b){
	if(!b){
		x = 1;
		y = 0;
		return;
	}
	exgcd(b,a%b);
	LL tmp = y;
	y = x - (a/b)*y;
	x = tmp;
}
LL lim1,lim2,lim3,lim4,liml,limr,T;
int main(){
	cin >> T;
	while(T --){
		scanf("%lld%lld%lld%lld%lld%lld%lld",&A,&B,&C,&P,&Q,&R,&S);
		if(!A && !B){
			if(C == 0){
				liml = max(P,R);
				limr = min(Q,S);
				if(limr < liml)printf("0\n");
				else printf("%lld\n",(Q - P + 1)*(S - R + 1));
			}
			else printf("0\n");
			continue;
		}
		G = Gcd(A,B);
		C *= -1;
		if(C%G){
			printf("0\n");
			continue;
		}
		a = A/G;b = B/G;c = C/G;
		exgcd(a,b);
		x *= c;y *= c;
		if(b < 0)b *= -1;
		if(a < 0)a *= -1;
		LL l = -100000000LL,r = 100000000LL;
		while(r - l > 1){
			LL mid = r + l >> 1;
			if(x + b*mid < P)l = mid;
			else r = mid;
		}
		lim1 = r;
		l = -100000000LL;r = 100000000LL;
		while(r - l > 1){
			LL mid = r + l >> 1;
			if(x + b*mid <= Q)l = mid;
			else r = mid;
		}
		lim2 = l;
		l = -100000000LL;r = 100000000LL;
		while(r - l > 1){
			LL mid = r + l >> 1;
			if(y + a*mid < R)l = mid;
			else r = mid;
		}
		lim3 = r;
		l = -100000000LL;r = 100000000LL;
		while(r - l > 1){
			LL mid = r + l >> 1;
			if(y + a*mid <= S)l = mid;
			else r = mid;
		}
		lim4 = l;
		if((A > 0) == (B > 0)){
			lim3 *= -1;
			lim4 *= -1;
			swap(lim3,lim4);
		}
		liml = max(lim1,lim3);
		limr = min(lim2,lim4);
		printf("%lld\n",max(0LL,limr - liml + 1));
		//cout << a << b << endl;
	}
}