http://codeforces.com/contest/627/problem/B

题意：有n个机器，机器非正常时每天生产b个物品，正常时每天生产a个物品，维修机器需要花费连续的k天(这k天什么都不能做)。有q次操作

1 a d，表示在d天有a个要求(每个要求需要一个物品)

2 p，表示从p天开始维修。(每一次维修是独立的)

机器初始是非正常的。

对每一个2操作，输出最多可以满足的要求数。

线段树，一个数组分别保存维修前和维修后能生产多少就ok了

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>

using namespace std;

#define N 201000
#define met(a, b) memset(a, b, sizeof(a))

const double PI = acos(-1.0);

typedef long long LL;

struct node
{
    int l, r, flag;
    LL num[4];
}tree[N*4];

LL a, b;

void Build (int rt, int l, int r)
{
    tree[rt].l = l, tree[rt].r = r;

    for (int i=0; i<2; i++)
        tree[rt].num[i] = 0;

    if (l==r) return;

    int mid = (l+r)/2;

    Build (rt<<1, l, mid);
    Build (rt<<1|1, mid+1, r);
}

void UP (int rt)
{
    for (int i=0; i<2; i++)
        tree[rt].num[i] = tree[rt<<1].num[i] + tree[rt<<1|1].num[i];
}

void Update (int rt, int k, LL e)
{
    if (tree[rt].l == tree[rt].r)
    {
        tree[rt].num[0] = min (tree[rt].num[0]+e, b);
        tree[rt].num[1] = min (tree[rt].num[1]+e, a);
        return;
    }

    int mid = (tree[rt].l+tree[rt].r)/2;

    if (k <= mid) Update (rt<<1, k, e);
    else Update (rt<<1|1, k, e);

    UP (rt);
}

LL Query (int rt, int l, int r, LL e)
{
    if (l > r) return 0;

    if (tree[rt].l==l && tree[rt].r==r)
        return tree[rt].num[e];

    int mid = (tree[rt].l+tree[rt].r)/2;

    if (r <= mid)
        return Query (rt<<1, l, r, e);
    else if (l > mid)
        return Query (rt<<1|1, l, r, e);
    else
        return Query (rt<<1, l, mid, e) + Query (rt<<1|1, mid+1, r, e);
}

int main ()
{
    int n, k, q;

    while (scanf ("%d %d %I64d %I64d %d", &n, &k, &a, &b, &q) != EOF)
    {
        Build (1, 1, n);

        while (q--)
        {
            LL x, y, z, ans1, ans2;
            scanf ("%I64d", &x);
            if (x==1)
            {
                scanf ("%I64d %I64d", &y, &z);
                Update (1, y, z);
            }
            else
            {
                scanf ("%I64d", &y);

                ans1 = Query (1, 1, y-1, 0);
                ans2 = Query (1, y+k, n, 1);

                printf ("%I64d\n", ans1+ans2);
            }
        }
    }
    return 0;
}