Codeforces 627B Factory Repairs 线段树

题目：627B - Factory Repairs

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.

题意：

有个工厂在n天内生产产品，本来机器是坏的，每天只能生产b件，修一次需要k天，修完后，每天可以生产a件，其中有q次询问，1开头，后面是d和ai，表示第d天有个ai的订单

2开头，后面有个d，表示这一天开始修理机器。每次2的时候，问工厂n天可以出售多少件产品？

分析：

线段树基础题型，1是单点增减，2是区间求和，sum二维数组分别维护修理前和修理后的生产效率，但是要注意都不可以超过订单额度，而且同一天可能有多个订单

区间求和的是修理前的天数和修理后的天数，相加即可

#include <bits/stdc++.h>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
int n,k,a,b;
const int N=2e5+1;
int sum[N<<2][2];
void pushup(int rt)
{
sum[rt][0]=sum[rt<<1][0]+sum[rt<<1|1][0];
sum[rt][1]=sum[rt<<1][1]+sum[rt<<1|1][1];
}
void update(int d,int ai,int l,int r,int rt)
{
if(d==l&&l==r){
sum[rt][0]=ai+sum[rt][0]>b?b:ai+sum[rt][0];
sum[rt][1]=ai+sum[rt][1]>a?a:ai+sum[rt][1];
return;
}
int m=(l+r)>>1;
if(d<=m)update(d,ai,lson);
else update(d,ai,rson);
pushup(rt);
}
int query(int L,int R,int id,int l,int r,int rt)
{
if(L<=l&&r<=R){
return sum[rt][id];
}
int ans=0;
int m=(l+r)>>1;
if(L<=m)ans+=query(L,R,id,lson);
if(m<R)ans+=query(L,R,id,rson);
return ans;
}
int main()
{
int q,op,d,ai;
//freopen("f.txt","r",stdin);
memset(sum,0,sizeof(sum));
scanf("%d%d%d%d%d",&n,&k,&a,&b,&q);
for(int i=0;i<q;i++){
scanf("%d",&op);
if(op==1){
scanf("%d%d",&d,&ai);
update(d,ai,1,n,1);
}
else{
scanf("%d",&d);
int ans=0;
if(d>1)ans=query(1,d-1,0,1,n,1);
// cout<<ans<<endl;
if(d+k<=n) ans+=query(d+k,n,1,1,n,1);
printf("%d\n",ans);
}
}
return 0;
}