洛谷P4891 序列

传送门

这题纯暴力竟然能过……

 //minamoto
 #include<cstdio>
 #include<iostream>
 #define mul(a,b) (1ll*a*b%P)
 using namespace std;
 #define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
 char buf[<<],*p1=buf,*p2=buf;
 int read(){
     #define num ch-'0'
     char ch;bool flag=;int res;
     while(!isdigit(ch=getc()))
     (ch=='-')&&(flag=true);
     for(res=num;isdigit(ch=getc());res=res*+num);
     (flag)&&(res=-res);
     #undef num
     return res;
 }
 char sr[<<],z[];int C=-,Z;
 inline void Ot(){fwrite(sr,,C+,stdout),C=-;}
 void print(int x){
     if(C><<)Ot();if(x<)sr[++C]=,x=-x;
     while(z[++Z]=x%+,x/=);
     while(sr[++C]=z[Z],--Z);sr[++C]='\n';
 }
 const int N=2e5+,P=1e9+;
 int n,m,a[N],b[N],c[N],ans=;
 int ksm(int a,int b){
     int res=;
     while(b){
         if(b&) res=mul(res,a);
         a=mul(a,a),b>>=;
     }
     return res;
 }
 int main(){
 //    freopen("testdata.in","r",stdin);
     n=read(),m=read();
     for(int i=;i<=n;++i)
     a[i]=read(),c[i]=max(c[i-],a[i]);
     for(int i=;i<=n;++i)
     b[i]=read(),ans=mul(min(b[i],c[i]),ans);
     while(m--){
         int op=read(),x=read(),y=read();
         if(op==){
             if(b[x]<c[x]){
                 ans=mul(ans,ksm(b[x],P-));
                 ans=mul(ans,min(y,c[x]));
             }
             b[x]=y;
         }else{
             a[x]=y;
             while(c[x]<y&&x<=n){
                 if(c[x]<b[x]){
                     ans=mul(ans,ksm(c[x],P-));
                     ans=mul(ans,min(y,b[x]));
                 }c[x++]=y;
             }
         }
         print(ans);
     }
     return Ot(),;
 }