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));
}
}
}

8VC Venture Cup 2016 - Final Round (Div. 2 Edition) D. Factory Repairs 树状数组的更多相关文章

  1. 8VC Venture Cup 2016 - Final Round (Div. 2 Edition)

    暴力 A - Orchestra import java.io.*; import java.util.*; public class Main { public static void main(S ...

  2. 8VC Venture Cup 2016 - Final Round (Div. 1 Edition) E - Preorder Test 树形dp

    E - Preorder Test 思路:想到二分答案了之后就不难啦, 对于每个答案用树形dp取check, 如果二分的值是val, dp[ i ]表示 i 这棵子树答案不低于val的可以访问的 最多 ...

  3. 8VC Venture Cup 2016 - Final Round (Div. 2 Edition) A

    A. Orchestra time limit per test 2 seconds memory limit per test 256 megabytes input standard input ...

  4. 8VC Venture Cup 2016 - Final Round (Div. 2 Edition) C. XOR Equation 数学

    C. XOR Equation 题目连接: http://www.codeforces.com/contest/635/problem/C Description Two positive integ ...

  5. 8VC Venture Cup 2016 - Final Round (Div. 2 Edition)B. sland Puzzle 水题

    B. sland Puzzle 题目连接: http://www.codeforces.com/contest/635/problem/B Description A remote island ch ...

  6. 8VC Venture Cup 2016 - Final Round (Div. 2 Edition) A. Orchestra 水题

    A. Orchestra 题目连接: http://www.codeforces.com/contest/635/problem/A Description Paul is at the orches ...

  7. Codeforces Round #393 (Div. 2) (8VC Venture Cup 2017 - Final Round Div. 2 Edition) E - Nikita and stack 线段树好题

    http://codeforces.com/contest/760/problem/E 题目大意:现在对栈有m个操作,但是顺序是乱的,现在每输入一个操作要求你输出当前的栈顶, 注意,已有操作要按它们的 ...

  8. Codeforces Round #393 (Div. 2) (8VC Venture Cup 2017 - Final Round Div. 2 Edition) D - Travel Card

    D - Travel Card 思路:dp,类似于单调队列优化. 其实可以写的更简单... #include<bits/stdc++.h> #define LL long long #de ...

  9. Codeforces Round #393 (Div. 2) (8VC Venture Cup 2017 - Final Round Div. 2 Edition)A 水 B 二分 C并查集

    A. Petr and a calendar time limit per test 2 seconds memory limit per test 256 megabytes input stand ...

随机推荐

  1. Python学习笔记 - day8 - 异常

    异常 在程序运行过程中,总会遇到各种各样的错误.有的错误是程序编写有问题造成的,比如本来应该输出整数结果输出了字符串,有的错误是用户输入造成的,比如让用户输入email地址,结果得到一个空字符串,这种 ...

  2. nvidia tk1使用记录--基本环境搭建

    前言 项目最开始是在X86+Nvidia(ubuntu+opencv+cuda)平台上实现,达到了期望性能,最近考虑将其移植到嵌入式平台,特别是最近nvidia出了tegra X1,基于和我们使用的g ...

  3. centos 引导盘

    # grub.conf generated by anaconda## Note that you do not have to rerun grub after making changes to ...

  4. mongodb-linux-x86_64

    卷 DataDisk 的文件夹 PATH 列表卷序列号为 4A8E-D95CD:.│  1.txt│  GNU-AGPL-3.0│  MPL-2│  README│  THIRD-PARTY-NOTI ...

  5. 105.Construct Binary Tree from Preorder and Inorder Traversal---《剑指offer》面试6

    题目链接 题目大意:根据先序遍历和中序遍历构造二叉树. 法一:DFS.根据模拟步骤,直接从先序和中序数组中找值然后加入二叉树中,即先从先序数组中确定根结点,然后再去中序数组中确定左子树和右子树的长度, ...

  6. 【openjudge】C15C Rabbit's Festival CDQ分治+并查集

    题目链接:http://poj.openjudge.cn/practice/C15C/ 题意:n 点 m 边 k 天.每条边在某一天会消失(仅仅那一天消失).问每一天有多少对点可以相互到达. 解法:开 ...

  7. 利用BeanUtils工具类封装表单数据

    一.BeanUtils工具类的使用 1.首先导入BeanUtils工具类的jar包 commons-beanutils-1.8.0.jar commons-logging-1.1.1.jar 2.se ...

  8. 取消div,a等标签点击效果

    当标签被设置onclick事件之后,在有些手机浏览器中,点击这些标签,会有点击变色效果.想要取消点击变色效果. 添加:div{-webkit-tap-highlight-color:rgba(0,0, ...

  9. Python 从基础------进阶------算法 系列

    1.简介                                                                                               关 ...

  10. 限制输入字符个数的jq插件

    (function($) { $.fn.extend( { limiter: function(limit, elem) { $(this).on("keyup focus", f ...