CF EDU 126 D - Progressions Covering

D - Progressions Covering

线段树维护差分序列

1. 把问题看作是每次给 b 数组的一个长度为 k 的子段减去一个等差数列，直到所有数小于等于 0
2. 因为子段长度严格为 k，因此对于当前 b 数组中最右边的大于 0 的元素，只能让它尽量作为子段右端点
3. 为了模拟这个过程，在子区间中减去等差数列，可认为是在差分序列中减去定值
4. 因此可以用线段树来模拟，需要给区间减定值（减等差数列）和求某个前缀和（求出当前最右边的元素值，看是否大于0）两个操作

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = 3e5 + 10;
ll a[N], b[N];
struct Node
{
	int l, r;
	ll add, sum;
}tr[N*4];
int n, k;

void pushup(int u)
{
	tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
}

void pushdown(int u)
{
	Node &root = tr[u], &left = tr[u << 1], &right = tr[u << 1 | 1];
	if (root.add)
	{
		left.add += root.add, right.add += root.add;
		left.sum += root.add * (left.r - left.l + 1);
		right.sum += root.add * (right.r - right.l + 1);
		root.add = 0;
	}
}

void build(int u, int l, int r)
{
	if (l == r)
	{
		tr[u] = {l, r, 0, a[l]};
		return;
	}
	tr[u] = {l, r, 0, 0};
	int mid = l + r >> 1;
	build(u << 1, l, mid);
	build(u << 1 | 1, mid + 1, r);
	pushup(u);
}

ll query(int u, int l, int r)
{
	if (tr[u].l >= l && tr[u].r <= r)
	{
		return tr[u].sum;
	}
	pushdown(u);
	ll sum = 0;
	int mid = tr[u].l + tr[u].r >> 1;
	if (l <= mid)
		sum += query(u << 1, l, r);
	if (r > mid)
		sum += query(u << 1 | 1, l, r);
	return sum;
}

void modify(int u, int l, int r, ll k)
{
	if (tr[u].l >= l && tr[u].r <= r)
	{
		tr[u].add += k;
		tr[u].sum += k * (tr[u].r - tr[u].l + 1);
		return;
	}
	pushdown(u);
	int mid = tr[u].l + tr[u].r >> 1;
	if (l <= mid)
		modify(u << 1, l, r, k);
	if (r > mid)
		modify(u << 1 | 1, l, r, k);
	pushup(u);
}

int main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> k;
	for (int i = 1; i <= n; i++)
		cin >> b[i];
	for (int i = 1; i <= n; i++)
		a[i] = b[i] - b[i-1];
	build(1, 1, n);
	ll ans = 0;
	for (int i = n; i >= 1; i--)
	{
		ll t = query(1, 1, i);
		if (t <= 0)
			continue;
		int r = max(i, k), l = r - k + 1;
		ll sub = i - l + 1;
		ll cnt = (t + sub - 1) / sub;
		modify(1, l, r, -cnt);
		ans += cnt;
	}
	cout << ans << endl;
	return 0;
}