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### 题解 CF1207F 【Remainder Problem】

posted on 2019-08-23 17:28:46 | under 题解 |

#include <bits/stdc++.h>
using namespace std;
int m,opt,x,y;
#define ll long long
ll ans[710][710],a[500005];
const int n=500000;
struct control
{
int ct,val;
control(int Ct,int Val=-1):ct(Ct),val(Val){}
inline control operator()(int Val)
{
return control(ct,Val);
}
}_endl(0),_prs(1),_setprecision(2);
struct FastIO
{
#define IOSIZE 1000000
char in[IOSIZE],*p,*pp,out[IOSIZE],*q,*qq,ch[20],*t,b,K,prs;
FastIO():p(in),pp(in),q(out),qq(out+IOSIZE),t(ch),b(1),K(6){}
~FastIO(){fwrite(out,1,q-out,stdout);}
inline char getch()
{
}
inline void putch(char x)
{
q==qq&&(fwrite(out,1,q-out,stdout),q=out),*q++=x;
}
inline void puts(const char str[]){fwrite(out,1,q-out,stdout),fwrite(str,1,strlen(str),stdout),q=out;}
inline void getline(string& s)
{
s="";
for(register char ch;(ch=getch())!='\n'&&b;)s+=ch;
}
#define indef(T) inline FastIO& operator>>(T& x)\
{\
x=0;register char f=0,ch;\
while(!isdigit(ch=getch())&&b)f|=ch=='-';\
while(isdigit(ch))x=(x<<1)+(x<<3)+(ch^48),ch=getch();\
return x=f?-x:x,*this;\
}
indef(int)
indef(long long)
inline FastIO& operator>>(char& ch){return ch=getch(),*this;}
inline FastIO& operator>>(string& s)
{
s="";register char ch;
while(isspace(ch=getch())&&b);
while(!isspace(ch)&&b)s+=ch,ch=getch();
return *this;
}
inline FastIO& operator>>(double& x)
{
x=0;register char f=0,ch;
double d=0.1;
while(!isdigit(ch=getch())&&b)f|=(ch=='-');
while(isdigit(ch))x=x*10+(ch^48),ch=getch();
if(ch=='.')while(isdigit(ch=getch()))x+=d*(ch^48),d*=0.1;
return x=f?-x:x,*this;
}
#define outdef(_T) inline FastIO& operator<<(_T x)\
{\
!x&&(putch('0'),0),x<0&&(putch('-'),x=-x);\
while(x)*t++=x%10+48,x/=10;\
while(t!=ch)*q++=*--t;\
return *this;\
}
outdef(int)
outdef(long long)
inline FastIO& operator<<(char ch){return putch(ch),*this;}
inline FastIO& operator<<(const char str[]){return puts(str),*this;}
inline FastIO& operator<<(const string& s){return puts(s.c_str()),*this;}
inline FastIO& operator<<(double x)
{
register int k=0;
this->operator<<(int(x));
putch('.');
x-=int(x);
prs&&(x+=5*pow(10,-K-1));
while(k<K)putch(int(x*=10)^48),x-=int(x),++k;
return *this;
}
inline FastIO& operator<<(const control& cl)
{
switch(cl.ct)
{
case 0:putch('\n');break;
case 1:prs=cl.val;break;
case 2:K=cl.val;break;
}
}
inline operator bool(){return b;}
}io;
int main(){
io>>m;
int sz=sqrt(n);//每个块的大小
while (m--){
io>>opt>>x>>y;
if (opt==2){
if (x<=sz)io<<ans[x][y]<<"\n";//可以直接进行的查询操作，直接输出
else{
ll sum=0;
for (int i=y;i<=n;i+=x)sum+=a[i];
io<<sum<<"\n";//否则暴力解决
}
}
else if (opt==1){
for (int p=1;p<=sz;p++)ans[p][x%p]+=y;//把修改x会产生影响的块的值都更新
a[x]+=y;//然后更新原数组
}
}
}