D. Factory Repairs
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 thedi-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.
Examples
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
output
3 6 4
input
5 4 10 1 6 1 1 5 1 5 5 1 3 2 1 5 2 2 1 2 2
output
7 1
Note
Consider the first sample.
We produce up to 1 thimble a day currently and will produce up to 2 thimbles a day after repairs. Repairs take 2 days.
For the first question, we are able to fill 1 order on day 1, no orders on days 2 and 3 since we are repairing, no orders on day 4 since no thimbles have been ordered for that day, and 2 orders for day 5 since we are limited to our production capacity, for a total of 3 orders filled.
For the third question, we are able to fill 1 order on day 1, 1 order on day 2, and 2 orders on day 5, for a total of 4 orders.
-------------------------------------------------EDITORIAL------------------------------------------------------------------------------
THIS PROBLEM CAN BE SOLVED EASILY IF WE MAINTAIN 2 DIFFERENT SEGMENT TREE, FIRST SEG TREE WILL CONTAIN ALL VALUES AT ALL INDEX NOT > a AND SECOND CONTAIN VALUES
NOT >b , NOT UPDATE BOTH TREE SUCH THAT EACH TREE INDEX WILL NOT CONTAIN VALUES > ITS RANGE a OR b ,
IN CASE OF QUERY , USE 1ST SEG TREE BEFORE REPAIR , AND SECOND SEG TREE AFTER REPAIR ....
------------------------------------------------CODE----------------------------------------------------------------------------------
#include<iostream>
#include<bits/stdc++.h>
typedef long long int lli;
#include<string.h>
long long int arr[1000005],seg1[10000005],seg2[10000005];
using namespace std;
#define inf 100000001
long long int qry1(int index,int start,int end,int qs,int qe)
{
if(start>end || end<qs || start>qe)
{
return 0;
}
if(start>=qs && end<=qe)
{
return seg1[index];
}
long long int q1=qry1(2*index,start,(start+end)/2,qs,qe);
long long int q2=qry1(2*index+1,((start+end)/2)+1,end,qs,qe);
return q1+q2;
}
void update1(int index,int start,int end,int ups,int upe,lli val,lli tt)
{
if(start>end || start>upe || end<ups) return ;
if(start>=ups && end<=upe)
{
seg1[index]+=val;
if(seg1[index]>tt) seg1[index]=tt;
return ;
}
update1(2*index,start,(start+end)/2,ups,upe,val,tt);
update1(2*index+1,((start+end)/2)+1,end,ups,upe,val,tt);
seg1[index]=seg1[2*index]+seg1[2*index+1];
}
long long int qry2(int index,int start,int end,int qs,int qe)
{
if(start>end || end<qs || start>qe)
{
return 0;
}
if(start>=qs && end<=qe)
{
return seg2[index];
}
long long int q1=qry2(2*index,start,(start+end)/2,qs,qe);
long long int q2=qry2(2*index+1,((start+end)/2)+1,end,qs,qe);
return q1+q2;
}
void update2(int index,int start,int end,int ups,int upe,lli val,lli tt)
{
if(start>end || start>upe || end<ups) return ;
if(start>=ups && end<=upe)
{
seg2[index]+=val;
if(seg2[index]>tt) seg2[index]=tt;
return ;
}
update2(2*index,start,(start+end)/2,ups,upe,val,tt);
update2(2*index+1,((start+end)/2)+1,end,ups,upe,val,tt);
seg2[index]=seg2[2*index]+seg2[2*index+1];
}
int main()
{
lli n,k,a,b,q;
cin>>n>>k>>a>>b>>q;
while(q--)
{
int type;
cin>>type;
if(type==1)
{
lli index,val;
cin>>index>>val;
index--;
update1(1,0,n-1,index,index,val,a);
update2(1,0,n-1,index,index,val,b);
}
else
{
long long int ans=0;
int index;
cin>>index;
index--;
ans=qry2(1,0,n-1,0,index-1);
ans+=qry1(1,0,n-1,index+k,n-1);
printf("%lld\n",ans);
}
}
}
.
? We use a max-segment tree which does these two operations: