洛谷P3396 哈希冲突（分块）

传送门

题解在此，讲的蛮清楚的->这里

我就贴个代码

 //minamoto
 #include<iostream>
 #include<cstdio>
 #include<cmath>
 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;
 inline 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=-;}
 inline 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';
 }
 inline char getop(){
     char ch;
     while((ch=getc())!='A'&&ch!='C');
     return ch;
 }
 const int N=,M=;
 int ans[M][M],v[N],s,n,m;
 int main(){
 //    freopen("testdata.in","r",stdin);
     n=read(),m=read(),s=pow(n,0.3333);
     for(int i=;i<=n;++i){
         v[i]=read();
         for(int j=;j<=s;++j)
         ans[j][i%j]+=v[i];
     }
     while(m--){
         char op=getop();int x=read(),y=read();
         switch(op){
             case 'A':{
                 if(x<=s) print(ans[x][y%x]);
                 else{
                     int res=;
                     for(int i=y;i<=n;i+=x) res+=v[i];
                     print(res);
                 }
                 break;
             }
             case 'C':{
                 for(int i=;i<=s;++i) ans[i][x%i]+=y-v[x];
                 v[x]=y;
                 break;
             }
         }
     }
     Ot();
     return ;
 }