JSOI2016灯塔跟这个一模一样（数据范围还比原题小）
题面：Luogu
60分题解：st表（也就是我上面说的那道题）
我们要对每个i求出
\[\max\left\{a_j+\left\lceil\sqrt{|i-j|}\right\rceil\right\}-a_i\]
\(\sqrt{|i-j|}\)只有\(\sqrt n\)种，直接枚举，用st表求出最大值
复杂度:\(O(n\sqrt n)\)

#include<bits/stdc++.h>
using namespace std;
inline void read(int& x)
{
    x = 0; char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
}
#define maxn 500005
int lg[maxn], st[maxn][21], h[maxn], n;
inline int getans(int l, int r)
{
    int tp = lg[r - l + 1];
    return max(st[l][tp], st[r - (1 << tp) + 1][tp]);
}
inline void st_init()
{
    for (int i = 2; i <= n; ++i) lg[i] = lg[i >> 1] + 1;
    for (int i = 1; i <= n; ++i) st[i][0] = h[i];
    for (int i = 1; i <= 19; ++i)
        for (int j = 1; j + (1 << i) - 1 <= n; ++j)
            st[j][i] = max(st[j][i - 1], st[j + (1 << (i - 1))][i - 1]);
}
int main()
{
    read(n);
    for (int i = 1; i <= n; ++i) read(h[i]);
    st_init();
    for (int i = 1; i <= n; ++i)
    {
        int ans = 0;
        for (int j = 1;; ++j)
        {
            int l = (j - 1) * (j - 1) + 1, r = j * j;
            if (i + l <= n) ans = max(ans, getans(i + l, min(n, i + r)) + j - h[i]);
            if (i - l >= 1) ans = max(ans, getans(max(1, i - r), i - l) + j - h[i]);
            if (i - r <= 1 && i + r >= n) break;
        }
        printf("%d\n", ans);
    }
    return 0;
}

---

还有一种用决策单调性+分治/单调队列的，复杂度\(O(n\log n)\)
首先是单调队列
单调队列维护形如\(\sqrt{x-a}+b\)的一系列图像
转移时二分出两条曲线相交的点

然后反过来再做一遍
因为上面的转移方程是
\[\max_{j=1}^{i}\left\{a_j+\left\lceil\sqrt{i-j}\right\rceil\right\}-a_i\]

#include<bits/stdc++.h>
using namespace std;

inline void read(int& x)
{
    x = 0; char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
}
#define maxn 500005
double sq[maxn];
int n, a[maxn], p[maxn], maxp[maxn];
struct Node
{
    int a, b;
    inline friend double calc(Node x, int X) { return sq[X - x.a] + x.b; }
    inline friend int get(Node X, Node Y)
    {
        int l = Y.a, r = n + 1, mid;
        while (l < r)
        {
            mid = (l + r) >> 1;
            if (calc(Y, mid) >= calc(X, mid)) r = mid;
            else l = mid + 1;
        }
        return l;
    }
};
template<typename T>
struct Deque
{
    int head, tail;
    T val[maxn];
    void clear() { head = 1, tail = 0; }
    T back() { return val[tail]; }
    T front() { return val[head]; }
    void pop_back() { --tail; }
    void pop_front() { ++head; }
    void push_back(T v) { val[++tail] = v; }
    void push_front(T v) { val[++head] = v; }
    bool empty() { return head > tail; }
};
Deque<Node> q;
Deque<int> Q;
void work()
{
    q.clear(); Q.clear();
    Node tp;
    for (int i = 1; i <= n; ++i)
    {
        tp = { i,a[i] };
        while (!q.empty() && get(q.back(), tp) <= Q.back()) q.pop_back(), Q.pop_back();
        Q.push_back(!q.empty() ? get(q.back(), tp) : 1);
        q.push_back(tp);
        while (Q.head < Q.tail && i >= Q.val[Q.head + 1]) ++Q.head, ++q.head;
        maxp[i] = max(maxp[i], (int)ceil(calc(q.front(), i)) - a[i]);
    }
}
int main()
{
    read(n);
    for (int i = 1; i <= n; ++i) read(a[i]), sq[i] = sqrt(i);
    work();
    reverse(a + 1, a + n + 1), reverse(maxp + 1, maxp + n + 1);
    work();
    for (int i = n; i; --i) printf("%d\n", maxp[i]);
    return 0;
}

---

分治
设\(p[l]\sim p[r]\)的最优决策点在\([L,R]\)间
设\(mid=\frac{l+r}{2}\)，扫一遍扫出来\(mid\)的最优决策点
然后就可以分治了（由决策单调性，\(p[l]\sim p[mid-1]\)d的最优决策点在\([L,pos]\)间）
分治又好写又跑得快，真香（总时间分治0.77s,单调队列2.41s(也可能是我单调队列写丑了，因为最优解也是单调队列）

#include<bits/stdc++.h>
using namespace std;

inline void read(int& x)
{
    x = 0; char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
}
#define maxn 500005
double sq[maxn], p[maxn];
int n, a[maxn];
void solve(int l, int r, int L, int R)
{
    if (l > r) return;
    int mid = (l + r) >> 1, pos = mid;
    double tmp, tmp_p;
    tmp_p = a[mid];
    for (int i = L; i <= min(R, mid); ++i)
    {
        tmp = a[i] + sq[mid - i];
        if (tmp > tmp_p) tmp_p = tmp, pos = i;
    }
    p[mid] = max(p[mid], tmp_p - a[mid]);
    solve(l, mid - 1, L, pos), solve(mid + 1, r, pos, R);
}

int main()
{
    read(n);
    for (int i = 1; i <= n; ++i) read(a[i]), sq[i] = sqrt(i);
    solve(1, n, 1, n);
    reverse(a + 1, a + n + 1), reverse(p + 1, p + n + 1);
    solve(1, n, 1, n);
    for (int i = n; i; --i) printf("%d\n", (int)ceil(p[i]));
    return 0;
}