【刷题】BZOJ 5418 [Noi2018]屠龙勇士

www.lydsy.com/JudgeOnline/upload/noi2018day2.pdf

Solution

将攻击的式子列出来，\(atk \times x-p \times y=a_i\)

这不就是个扩欧的裸式子嘛，求出 \(x\) 的解的式子 \(x=x_0+r \times dis\)，其中 \(r\) 为任意整数，\(dis\) 为不定方程解的间隔

上面的式子发现又是一个同余方程

对于每一条龙都是一个同余方程，那么就是要解一个同余方程组的最小解

用扩展CRT就好了

#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
const int MAXN=100000+10;
ll T,n,m,h[MAXN],p[MAXN],rem[MAXN],mod[MAXN],v[MAXN],k[MAXN],lm;
std::multiset<ll> S;
template<typename T> inline void read(T &x)
{
	T data=0,w=1;
	char ch=0;
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
	x=data*w;
}
template<typename T> inline void write(T x,char ch='\0')
{
	if(x<0)putchar('-'),x=-x;
	if(x>9)write(x/10);
	putchar(x%10+'0');
	if(ch!='\0')putchar(ch);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
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;}
inline ll mul(ll a,ll b,ll Mod)
{
	ll res=0;
	if(b<0)a=(a%Mod+Mod)%Mod,b=(b%Mod+Mod)%Mod;
	while(b)
	{
		if(b&1)res=(res+a)%Mod;
		a=(a+a)%Mod;
		b>>=1;
	}
	return res;
}
inline ll exgcd(ll a,ll b,ll &x,ll &y)
{
	if(b==0)
	{
		x=1;
		y=0;
		return a;
	}
	ll r=exgcd(b,a%b,x,y);
	ll t=x;
	x=y;
	y=t-(a/b)*y;
	return r;
}
inline ll exCRT()
{
	ll M=mod[1],A=rem[1],t,d,x,y;
	for(register ll i=2;i<=n;++i)
	{
		d=exgcd(M,mod[i],x,y);
		if((rem[i]-A)%d)return -1;
		t=mod[i]/d;
		x=(mul(x,(rem[i]-A)/d,t)%t+t)%t;
		ll tmp=M;M=M/d*mod[i];
		A=(mul(tmp,x,M)+A)%M;;
	}
	A=(A%M+M)%M;
	while(A<lm)A+=M;
	return A;
}
inline void solve()
{
	lm=0;
	memset(rem,0,sizeof(rem));
	memset(mod,0,sizeof(mod));
	for(register int i=1;i<=n;++i)
	{
		if(*S.begin()>=h[i])k[i]=*S.begin(),S.erase(S.begin());
		else
		{
			std::multiset<ll>::iterator it=--S.upper_bound(h[i]);
			k[i]=*it,S.erase(it);
		}
		S.insert(v[i]);
	}
	bool mk=1;
	for(register int i=1;i<=n;++i)
	{
		ll x,y,a=k[i],b=p[i],c=h[i],d=exgcd(a,b,x,y);
		if(p[i]==1)chkmax(lm,(h[i]+k[i]-1)/k[i]);
		else mk=0;
		if(c%d)
		{
			puts("-1");
			return ;
		}
		mod[i]=p[i]/d,rem[i]=mul(x,c/d,mod[i]);
		if(!rem[i])chkmax(lm,mod[i]);
	}
	if(mk)write(lm,'\n');
	else write(exCRT(),'\n');
}
int main()
{
	read(T);
	while(T--)
	{
		read(n);read(m);S.clear();
		for(register int i=1;i<=n;++i)read(h[i]);
		for(register int i=1;i<=n;++i)read(p[i]);
		for(register int i=1;i<=n;++i)read(v[i]);
		for(register int i=1,x;i<=m;++i)read(x),S.insert(x);
		solve();
	}
	return 0;
}