[NOI2018]屠龍勇士

description

題面

solution

主要算法是擴展中國剩餘定理，但細節較多

快速乘

il ll upd(ll a,ll b,ll mod){return ((a+b)%mod+mod)%mod;}
il ll mul(ll a,ll b,ll mod){
	RG ll ret=0;
	if(b<0)a=-a,b=-b;
	for(;b;b>>=1,a=upd(a,a,mod))
		if(b&1)ret=upd(ret,a,mod);
	return ret;
}

求同餘方程的通解

須要注意同餘方程$ax\equiv b(mod\ p)$的通解 求出一個特解$x_0$後，應該有$x=x_0\pm\frac{p}{g}$ 所以最後解出的同餘方程應該是$x\equiv x_0(mod\ \frac{p}{g})$

il ll solvemod(ll a,ll b,ll &p,ll &r,ll &P){
	ll x,y,g;
	a=(a%p+p)%p;b=(b%p+p)%p;
	Exgcd(a,p,x,y,g);
	if(b%g)return -1;
	x=(x%p+p)%p;P=p/g;
	x=mul(b/g,x,P);
	r=x;return 0;
}

擴展CRT

il ll exchina(ll *a,ll *b,ll *p,ll n,ll &L){
	ll r1,lcm1,ri,lcmi,x,y,g;
	if(solvemod(a[1],b[1],p[1],r1,lcm1)==-1)return -1;
	for(RG int i=2;i<=n;i++){
		if(solvemod(a[i],b[i],p[i],ri,lcmi)==-1)return -1;
		Exgcd(lcm1,lcmi,x,y,g);
		if((ri-r1)%g)return -1;
		L=lcm(lcm1,lcmi);
		r1=upd(r1,mul(mul((ri-r1)/g,x,L),lcm1,L),L);
		lcm1=L;
	}
	return r1;
}

關於multiset中S.erase()的使用...

若是傳的參是一個迭代器，那麼只會刪除當前迭代器對應的鍵值， 若是傳的是一個數，那麼set會刪除裏面全部的這個數...

code

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
#define RG register
#define il inline
#define FILE "dragon"
using namespace std;
typedef long long ll;
typedef double dd;
const int N=100010;
const int inf=2147483647;
const ll INF=1e18+1;
il void file(){
	freopen(FILE".in","r",stdin);
	freopen(FILE".out","w",stdout);
}
il ll read(){
	RG ll d=0,w=1;char ch=getchar();
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch<='9'&&ch>='0')d=d*10+ch-48,ch=getchar();	
	return d*w;
}

//Case set
multiset<ll>S;
multiset<ll>::iterator tmp,pre;
il void geta(ll *b,ll *s,ll *k,int n,int m,ll *a){
	S.clear();
	for(RG int i=1;i<=m;i++)S.insert(s[i]);
	for(RG int i=1;i<=n;i++){
		tmp=S.begin();//printf("%lld,%lld\n",*tmp,b[i]);
		if((*tmp)>b[i]){a[i]=*tmp;S.erase(tmp);}
		else{
			tmp=S.upper_bound(b[i]);tmp--;
			a[i]=*tmp;S.erase(tmp);
		}
		S.insert(k[i]);
	}
}

//math
il ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
il ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
il void Exgcd(ll a,ll b,ll &x,ll &y,ll &g){
	if(!b){g=a;x=1;y=0;return;}
	Exgcd(b,a%b,y,x,g);y-=a/b*x;
}
il ll upd(ll a,ll b,ll mod){return ((a+b)%mod+mod)%mod;}
il ll mul(ll a,ll b,ll mod){
	RG ll ret=0;
	if(b<0)a=-a,b=-b;
	for(;b;b>>=1,a=upd(a,a,mod))
		if(b&1)ret=upd(ret,a,mod);
	return ret;
}

il ll solvemod(ll a,ll b,ll &p,ll &r,ll &P){
	ll x,y,g;
	a=(a%p+p)%p;b=(b%p+p)%p;
	Exgcd(a,p,x,y,g);
	if(b%g)return -1;
	x=(x%p+p)%p;P=p/g;
	x=mul(b/g,x,P);
	r=x;
	return 0;
}

il ll exchina(ll *a,ll *b,ll *p,ll n,ll &L){
	ll r1,lcm1,ri,lcmi,x,y,g;
	if(solvemod(a[1],b[1],p[1],r1,lcm1)==-1)return -1;
	for(RG int i=2;i<=n;i++){
		if(solvemod(a[i],b[i],p[i],ri,lcmi)==-1)return -1;
		Exgcd(lcm1,lcmi,x,y,g);
		if((ri-r1)%g)return -1;
		L=lcm(lcm1,lcmi);
		r1=upd(r1,mul(mul((ri-r1)/g,x,L),lcm1,L),L);
		lcm1=L;
	}
	return r1;
}

//input
int n,m;ll a[N],b[N],p[N],s[N],k[N],ans,ret,tot;
il void work(){
	n=read();m=read();
	for(RG int i=1;i<=n;i++)b[i]=read();
	for(RG int i=1;i<=n;i++)p[i]=read();
	for(RG int i=1;i<=n;i++)k[i]=read();
	for(RG int i=1;i<=m;i++)s[i]=read();
	geta(b,s,k,n,m,a);
	ans=exchina(a,b,p,n,tot);
	if(ans==-1){puts("-1");return;}
	
	ret=0;for(RG int i=1;i<=n;i++)ret=max(ret,(b[i]+a[i]-1)/a[i]);
	ans=ans+ret/tot*tot+(ret%tot>ans)*tot;
	printf("%lld\n",ans);
}

int main()
{	
	RG int T=read();
	while(T--)
		work();
	return 0;
}