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[1<<21],*p1=buf,*p2=buf;
inline int read(){
    #define num ch-'0'
    char ch;bool flag=0;int res;
    while(!isdigit(ch=getc()))
    (ch=='-')&&(flag=true);
    for(res=num;isdigit(ch=getc());res=res*10+num);
    (flag)&&(res=-res);
    #undef num
    return res;
}
char sr[1<<21],z[20];int C=-1,Z;
inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
inline void print(int x){
    if(C>1<<20)Ot();if(x<0)sr[++C]=45,x=-x;
    while(z[++Z]=x%10+48,x/=10);
    while(sr[++C]=z[Z],--Z);sr[++C]='
';
}
inline char getop(){
    char ch;
    while((ch=getc())!='A'&&ch!='C');
    return ch;
}
const int N=150005,M=2005;
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=1;i<=n;++i){
        v[i]=read();
        for(int j=1;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=0;
                    for(int i=y;i<=n;i+=x) res+=v[i];
                    print(res);
                }
                break;
            }
            case 'C':{
                for(int i=1;i<=s;++i) ans[i][x%i]+=y-v[x];
                v[x]=y;
                break;
            }
        }
    }
    Ot();
    return 0;
}