8VC Venture Cup 2016 - Final Round (Div. 2 Edition) D. Factory Repairs 树状数组

D. Factory Repairs

题目连接：

http://www.codeforces.com/contest/635/problem/D

Description

A factory produces thimbles in bulk. Typically, it can produce up to a thimbles a day. However, some of the machinery is defective, so it can currently only produce b thimbles each day. The factory intends to choose a k-day period to do maintenance and construction; it cannot produce any thimbles during this time, but will be restored to its full production of a thimbles per day after the k days are complete.

Initially, no orders are pending. The factory receives updates of the form di, ai, indicating that ai new orders have been placed for the di-th day. Each order requires a single thimble to be produced on precisely the specified day. The factory may opt to fill as many or as few of the orders in a single batch as it likes.

As orders come in, the factory owner would like to know the maximum number of orders he will be able to fill if he starts repairs on a given day pi. Help the owner answer his questions.

Input

The first line contains five integers n, k, a, b, and q (1 ≤ k ≤ n ≤ 200 000, 1 ≤ b < a ≤ 10 000, 1 ≤ q ≤ 200 000) — the number of days, the length of the repair time, the production rates of the factory, and the number of updates, respectively.

The next q lines contain the descriptions of the queries. Each query is of one of the following two forms:

1 di ai (1 ≤ di ≤ n, 1 ≤ ai ≤ 10 000), representing an update of ai orders on day di, or
2 pi (1 ≤ pi ≤ n - k + 1), representing a question: at the moment, how many orders could be filled if the factory decided to commence repairs on day pi?
It's guaranteed that the input will contain at least one query of the second type.

Output

For each query of the second type, print a line containing a single integer — the maximum number of orders that the factory can fill over all n days.

Sample Input

5 2 2 1 8
1 1 2
1 5 3
1 2 1
2 2
1 4 2
1 3 2
2 1
2 3

Sample Output

3
6
4

Hint

题意

最多n天，然后你每天普通情况下可以产生b个东西，好的情况下可以产生b个东西

由普通的情况到好的情况需要k天的休息，就这k天啥也干不了

q次询问

现在你有两个操作

1 x y，第x天需要y个东西（不独立）

2 x 在第x天进行休息，然后问你最多能够满足多少个东西的需求（独立的）

题解：

树状数组就好了

k天以前的，我就每个点的最大值就是b

k天以后，每个点的最大值是a

然后维护一下区间和就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn =  2e5+7;
int n,k,a,b,q;
int c[maxn][2];
int A[maxn];
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int y,int z)
{
    for(int i=x;i<maxn;i+=lowbit(i))
        c[i][z]+=y;
}
int getsum(int x,int z)
{
    int ans = 0;
    for(int i=x;i;i-=lowbit(i))
        ans+=c[i][z];
    return ans;
}
int main()
{
    scanf("%d%d%d%d%d",&n,&k,&a,&b,&q);
    for(int i=1;i<=q;i++)
    {
        int op;scanf("%d",&op);
        if(op==1)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            int p1 = A[x];A[x]+=y;
            update(x,min(b,A[x])-min(b,p1),0);
            update(x,min(a,A[x])-min(a,p1),1);
        }
        else
        {
            int x;scanf("%d",&x);
            printf("%d\n",getsum(x-1,0)+getsum(n,1)-getsum(x+k-1,1));
        }
    }
}