題目大意

n個房間各有一些需要更換的檯燈。每個月買m盞，包括上月累積，從左到右對各房間換燈泡，如果房間裡需要更換的數量大於手中的燈泡數則跳過。詢問某月累計對多少房間更換燈泡，以及剩下的燈泡數。

思路

線段樹維護區間最小值，每次查詢左邊第一個小於手中燈泡數的房間。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define INF 0x3f3f3f3f
#define rep0(i, n) for (int i = 0; i < n; i++)
#define rep1(i, n) for (int i = 1; i <= n; i++)
#define rep_0(i, n) for (int i = n - 1; i >= 0; i--)
#define rep_1(i, n) for (int i = n; i > 0; i--)
#define MAX(x, y) (((x) > (y)) ? (x) : (y))
#define MIN(x, y) (((x) < (y)) ? (x) : (y))
#define mem(x, y) memset(x, y, sizeof(x))
#define MAXN 100010
using namespace std;
struct Node
{
    int mn;

} tree[MAXN << 2];
int ans[MAXN], rm[MAXN], qu[MAXN];
int a[MAXN];
void push_up(int rt, int l, int r)
{
    if (l == r)
        return;
    tree[rt].mn = MIN(tree[rt << 1].mn, tree[rt << 1 | 1].mn);
}
void build(int rt, int l, int r)
{
    if (l == r)
    {
        tree[rt].mn = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(rt << 1, l, mid);
    build(rt << 1 | 1, mid + 1, r);

    push_up(rt, l, r);
 

}
int update(int rt, int l, int r, int val)
{
    if (tree[rt].mn > val)
        return 0;

    if (l == r)
    {
        int tmp = tree[rt].mn;
        tree[rt].mn = INF;

        return tmp;
    }
    int mid = (l + r) >> 1;
    int ans = 0;

    if (tree[rt << 1].mn <= val)
        ans = update(rt << 1, l, mid, val);
    else if (tree[rt << 1 | 1].mn <= val)
        ans = update(rt << 1 | 1, mid + 1, r, val);

    push_up(rt, l, r);

    return ans;

}


int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
    #endif // ONLINE_JUDGE

    int n, m, q, mx_mon = 0;
    scanf("%d %d", &n, &m);

    for (int i = 1; i <= n; i++)
        scanf("%d", a + i);

    build(1, 1, n);

    scanf("%d", &q);

    for (int i = 1; i <= q; i++)
    {
        scanf("%d", qu + i);
        mx_mon = MAX(mx_mon, qu[i]);
    }

    int cnt = n, lmp = 0, mon = 1;
    for (; mon <= mx_mon; mon++)
    {
        if (!cnt)
        {
            ans[mon] = n - cnt;
            rm[mon] = lmp;
            continue;
        }
        int tmp;
        lmp += m;
        while (cnt && (tmp = update(1, 1, n, lmp)))
        {
            cnt--;
            lmp -= tmp;
        }
        ans[mon] = n - cnt;
        rm[mon] = lmp;
    }



    for (int i = 1; i <= q; i++)
        printf("%d %d\n", ans[qu[i]], rm[qu[i]]);



    return 0;
}