[NOI2018]屠龙勇士

题目链接：BZOJ，[洛谷](https://www.luogu.org/problemnew/show/P47740)

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对于每条龙，我们可以算出攻击次数\(x_i\)，满足方程\(x_i\equiv Z_i(\%R_i)\)，\(R_i=\dfrac{p_i}{\gcd(p_i,ATK_i)}\)，\(Z_i\)则为首次将巨龙杀死的攻击次数（\(Z_i\)可能大于\(R_i\)，但是不能\(\%\)掉）

然后我们就可以用ex_crt将所有方程合并起来

最后求出的解\(X\)可能\(<\max\limits_{i=1}^m\{Z_i\}\)，那么我们就把\(X\)不断加\(lcm(R_1,R_2,...R_m)\)，直到满足即可

/*program from Wolfycz*/
#include<set>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
typedef long long ll;
typedef long double ld;
typedef unsigned int ui;
typedef std::multiset<ll> msi;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
	int f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')    f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
template<typename T>inline T read(T x){
	int f=1;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)    putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
template<typename T>inline T min(T x,T y){return x<y?x:y;}
template<typename T>inline T max(T x,T y){return x>y?x:y;}
template<typename T>inline T swap(T &x,T &y){T t=x; x=y,y=t;}
const int N=1e5;
namespace Math{
	ll mul(ll _a,ll _b,ll _p){
		ll _c=(ld)_a*_b/_p+0.5;
		ll _ans=_a*_b-_c*_p;
		if (_ans<0)	_ans+=_p;
		return _ans;
	}
	ll gcd(ll a,ll b){return !b?a:gcd(b,a%b);}
	void exgcd(ll a,ll b,ll &x,ll &y){
		if (!b){x=1,y=0;return;}
		exgcd(b,a%b,x,y);
		ll t=x; x=y,y=t-a/b*y;
	}
	ll Ex_GCD(ll a,ll b,ll c){
		ll d=gcd(a,b),x,y;
		if (c%d)	return -1;
		a/=d,b/=d,c/=d;
		exgcd(a,b,x,y);
		x=(mul(x,c,b)+b)%b;
		if ((c-a*x)/b>0)	x+=b;
		return x;
	}
	ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
}
msi Sw;//Sword
ll h[N+10],Z[N+10],p[N+10],R[N+10];
int V[N+10];
// x = Zi ( % Ri )
using namespace Math;
int main(){
	for (int T=read(0);T;T--){
		Sw.clear();
		int n=read(0),m=read(0); ll Max=0;
		for (int i=1;i<=n;i++)	h[i]=read(0ll);
		for (int i=1;i<=n;i++)	p[i]=read(0ll);
		for (int i=1;i<=n;i++)	V[i]=read(0);
		for (int i=1;i<=m;i++)	Sw.insert(read(0));
		bool Wrong=0;
		for (int i=1;i<=n;i++){
			msi::iterator it=Sw.upper_bound(h[i]);
			if (it!=Sw.begin())	it--;
			Z[i]=h[i]/(*it),h[i]%=*it;
			ll tmp=Ex_GCD(*it,p[i],h[i]);
			if (tmp==-1){Wrong=1;break;}
			Z[i]+=tmp,R[i]=p[i]/Math::gcd(*it,p[i]);
			Sw.erase(it),Sw.insert(V[i]);
			Max=max(Max,Z[i]);
		}
		if (Wrong){
			printf("-1\n");
			continue;
		}
		ll Z1=Z[1],R1=R[1];
		for (int i=2;i<=n;i++){
			ll tmp=Ex_GCD(R1,R[i],Z[i]-Z1);
			if (tmp==-1){
				Wrong=1;
				break;
			}
			ll Lcm=lcm(R1,R[i]);
			Z1=(Z1+mul(tmp,R1,Lcm))%Lcm,R1=Lcm;
		}
		if (Wrong){
			printf("-1\n");
			continue;
		}
		ll Ans=Ex_GCD(1,R1,Z1);
		if (Ans<Max)	Ans+=R1*((Max-Ans-1)/R1+1);
		printf("%lld\n",Ans);
	}
	return 0;
}