Remainder Problem
题面:
You are given an array a consisting of 500000 integers (numbered from 1 to 500000). Initially all elements of a are zero.
You have to process two types of queries to this array:
1 x y — increase ax by y;
2 x y — compute ∑i∈R(x,y)ai, where R(x,y) is the set of all integers from 1 to 500000 which have remainder y modulo x.
Can you process all the queries?
题意:
给你一个长度500000的数组，有两个操作
1 x y 将a[ x ] +=y;
2 x y 将所有下标%x==y的位置所在的数相加并输出
思路: 每次将模数小于720的预处理，若是所询问模数大于sqrt(500000)直接暴力统计

#include<bits/stdc++.h>
using namespace std;
const int maxl=500010;
int q,n=maxl-10,len,cnt;
long long ans;
int a[maxl];
long long val[720][720]; 
int main()
{
	len=sqrt(n);
	cnt=cnt/len+1;
	scanf("%d",&q);
	int op,x,y;
	for(int i=1;i<=q;i++)
	{
		scanf("%d%d%d",&op,&x,&y);
		if(op==1)
		{
			a[x]+=y;
			for(int i=1;i<720;i++)
				val[i][x%i]+=y;
		}
		else
		{
			ans=0;
			if(x<720)
				ans=val[x][y];
			else
			{
				for(int i=0;i*x+y<=n;i++)
					ans+=a[i*x+y];
			}
			printf("%lld\n",ans);	
		}
	}
	return 0;	
}