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 d_i a_i (1 ≤ d_i ≤ n,1 ≤ a_i ≤ 10 000), representing an update ofa_i orders on dayd_i, or
• 2 p_i (1 ≤ p_i ≤ n - k + 1), representing a question: at the moment, how many orders could be filled if the factory decided to commence repairs on day p_i?

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 alln 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|1using 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;}