这题到底是什么东西……数学题太珂怕了……
1 //minamoto 2 #include<iostream> 3 #include<cstdio> 4 #include<algorithm> 5 #include<set> 6 #define ll long long 7 #define GG {puts("-1");return;} 8 using namespace std; 9 const int N=1e5+5; 10 #define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) 11 char buf[1<<21],*p1=buf,*p2=buf; 12 template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;} 13 inline ll read(){ 14 #define num ch-'0' 15 char ch;bool flag=0;ll res; 16 while((ch=getc())>'9'||ch<'0') 17 (ch=='-')&&(flag=true); 18 for(res=num;(ch=getc())<='9'&&ch>='0';res=res*10+num); 19 (flag)&&(res=-res); 20 #undef num 21 return res; 22 } 23 multiset<ll> s;int n,m;ll mod[N],xs[N],cs[N],st[N],gif[N];int T; 24 inline void exgcd(ll a,ll &x,ll b,ll &y){ 25 if(!b) return (void)(x=1,y=0); 26 exgcd(b,x,a%b,y);ll t=x;x=y,y=t-(a/b)*y; 27 } 28 inline ll gcd(ll a,ll b){ 29 if(a<b) swap(a,b); 30 while(b^=a^=b^=a%=b);return a; 31 } 32 inline ll inv(ll a,ll p){a%=p;ll x,y;exgcd(a,x,p,y);return x<0?x+p:x;} 33 inline ll mul(ll a,ll b,const ll &mod){ 34 ll res=0;b%=mod,b=b>0?b:b+mod; 35 for(;b;b>>=1,a=(a+a)%mod) (res+=a*(b&1))%=mod; 36 return res; 37 } 38 inline void spj(){ 39 ll res=0; 40 for(int i=1;i<=n;++i) cmax(res,(cs[i]+xs[i]-1)/xs[i]); 41 printf("%lld ",res); 42 } 43 inline void spj2(){ 44 ll lc=1; 45 for(int i=1;i<=n;++i){ 46 ll ds=mod[i]/gcd(mod[i],xs[i]); 47 lc=lc/gcd(lc,ds)*ds; 48 } 49 printf("%lld ",lc); 50 } 51 void solve(){ 52 n=read(),m=read(); 53 for(int i=1;i<=n;++i) cs[i]=read(); 54 for(int i=1;i<=n;++i) mod[i]=read(); 55 for(int i=1;i<=n;++i) gif[i]=read(); 56 for(int i=1;i<=m;++i) st[i]=read(); 57 for(int i=1;i<=m;++i) s.insert(st[i]); 58 multiset<ll>::iterator ii; 59 for(int i=1;i<=n;++i){ 60 ii=(cs[i]<(*s.begin()))?s.begin():(--s.upper_bound(cs[i])); 61 xs[i]=*ii,s.erase(ii),s.insert(gif[i]); 62 } 63 for(int i=1;i<=m;++i) 64 if(mod[i]==cs[i]){spj2();return;} 65 for(int i=1;i<=n;++i) 66 if(mod[i]==1){spj();return;} 67 for(int i=1;i<=n;++i) xs[i]%=mod[i]; 68 for(int i=1;i<=n;++i) 69 if(xs[i]==0){ 70 if(mod[i]==cs[i]) xs[i]=1,mod[i]=1,cs[i]=0; 71 else GG; 72 } 73 for(int i=1;i<=n;++i){ 74 ll sx,sy,g=gcd(xs[i],mod[i]); 75 if(cs[i]%g!=0) GG; 76 exgcd(xs[i],sx,mod[i],sy); 77 mod[i]=mod[i]/g;sx=(sx%mod[i]+mod[i])%mod[i],cs[i]=mul(sx,cs[i]/g,mod[i]); 78 } 79 for(int i=1;i<n;++i){ 80 ll g=gcd(mod[i],mod[i+1]);if((cs[i+1]-cs[i])%g!=0) GG; 81 ll ncs=mul(inv(mod[i]/g,mod[i+1]/g),(cs[i+1]-cs[i])/g,mod[i+1]/g); 82 ll nmod=mod[i]/g*mod[i+1]; 83 ncs=mul(ncs,mod[i],nmod),(ncs+=cs[i])%=nmod,cs[i+1]=ncs,mod[i+1]=nmod; 84 }printf("%lld ",cs[n]); 85 } 86 inline void clear(){s.clear();} 87 int main(){ 88 // freopen("testdata.in","r",stdin); 89 T=read(); 90 for(int i=1;i<=T;++i) solve(),clear(); 91 return 0; 92 }