题解在此，讲的蛮清楚的->这里
我就贴个代码
1 //minamoto 2 #include<iostream> 3 #include<cstdio> 4 #include<cmath> 5 using namespace std; 6 #define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) 7 char buf[1<<21],*p1=buf,*p2=buf; 8 inline int read(){ 9 #define num ch-'0' 10 char ch;bool flag=0;int res; 11 while(!isdigit(ch=getc())) 12 (ch=='-')&&(flag=true); 13 for(res=num;isdigit(ch=getc());res=res*10+num); 14 (flag)&&(res=-res); 15 #undef num 16 return res; 17 } 18 char sr[1<<21],z[20];int C=-1,Z; 19 inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;} 20 inline void print(int x){ 21 if(C>1<<20)Ot();if(x<0)sr[++C]=45,x=-x; 22 while(z[++Z]=x%10+48,x/=10); 23 while(sr[++C]=z[Z],--Z);sr[++C]=' '; 24 } 25 inline char getop(){ 26 char ch; 27 while((ch=getc())!='A'&&ch!='C'); 28 return ch; 29 } 30 const int N=150005,M=2005; 31 int ans[M][M],v[N],s,n,m; 32 int main(){ 33 // freopen("testdata.in","r",stdin); 34 n=read(),m=read(),s=pow(n,0.3333); 35 for(int i=1;i<=n;++i){ 36 v[i]=read(); 37 for(int j=1;j<=s;++j) 38 ans[j][i%j]+=v[i]; 39 } 40 while(m--){ 41 char op=getop();int x=read(),y=read(); 42 switch(op){ 43 case 'A':{ 44 if(x<=s) print(ans[x][y%x]); 45 else{ 46 int res=0; 47 for(int i=y;i<=n;i+=x) res+=v[i]; 48 print(res); 49 } 50 break; 51 } 52 case 'C':{ 53 for(int i=1;i<=s;++i) ans[i][x%i]+=y-v[x]; 54 v[x]=y; 55 break; 56 } 57 } 58 } 59 Ot(); 60 return 0; 61 }