[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;
}
}