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(),;

 }