F Remainder Problem

time limit per test 4 seconds
memory limit per test 512 megabytes
inputstandard input
outputstandard output
You are given an array a consisting of 500000integers (numbered from 11 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?

Input
The first line contains one integer q (1≤q≤500000) — the number of queries.

Then q lines follow, each describing a query. The i-th line contains three integers ti, xi and yi (1≤ti≤2). If ti=1, then it is a query of the first type, 1≤xi≤500000, and −1000≤yi≤10000. If ti=2, then it it a query of the second type, 1≤xi≤500000, and 0≤yi<xi.

It is guaranteed that there will be at least one query of type 22.

Output
For each query of type 2 print one integer — the answer to it.

Example
input
5
1 3 4
2 3 0
2 4 3
1 4 -4
2 1 0
output
4
4
0

题意：此题涉及两个操作:当为1时，输入两个数x和y，就是将a[x]+y;当为2时，输入两个数x和y，要求满足x%i==y的a[i]的和为多少。

思路：由于查询次数最大为5e5次，如果单纯的查询的话肯定会超时，如果单纯的建立一个记忆化数组的话x为50万y为1000肯定会超内存，所以我们记忆化到根号下五十万之前（得出根号下50万之前的所有结果），对于之后的结果直接暴力求解。（算出来时间刚好卡在3800多ms）

PS:应定义为longlong类型

AC代码如下：

#include<iostream>
#include<cstring>
using namespace std;
const int maxn = 5e5 + 5;
//1表示a[x]+y
//2表示返回满足x%i==y的a[i]的和
typedef long long ll;
ll a[maxn];
ll a1[705][705];
int main()
	{
	memset(a, 0, sizeof(a));
	memset(a1, 0, sizeof(a1));
	int t;
	cin >> t;
	while (t--)
	{
		int flag,x, y;
		cin >> flag >> x >> y;
		if (flag == 1)
		{
			a[x] += y;
			for (int i = 1; i <= 700; i++)
				a1[i][x%i] += y;

		}
		if (flag == 2)
		{

			if (x <= 700)cout << a1[x][y] << endl;
			else{
				ll ans = 0;
				//注意i每次加x
				for (int i = y; i <= 500000; i += x)
					ans += a[i];
				cout << ans << endl;			
			}
		}
	}
	return 0;
}