www.lydsy.com/JudgeOnline/upload/noi2018day2.pdf
Solution
将攻击的式子列出来，(atk imes x-p imes 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 imes 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=' ')
{
if(x<0)putchar('-'),x=-x;
if(x>9)write(x/10);
putchar(x%10+'0');
if(ch!=' ')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,'
');
else write(exCRT(),'
');
}
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;
}