D. Running Miles

D. Running Miles

There is a street with $n$ sights, with sight number $i$ being $i$ miles from the beginning of the street. Sight number $i$ has beauty $b_i$. You want to start your morning jog $l$ miles and end it $r$ miles from the beginning of the street. By the time you run, you will see sights you run by (including sights at $l$ and $r$ miles from the start). You are interested in the $3$ most beautiful sights along your jog, but every mile you run, you get more and more tired.

So choose $l$ and $r$, such that there are at least $3$ sights you run by, and the sum of beauties of the $3$ most beautiful sights minus the distance in miles you have to run is maximized. More formally, choose $l$ and $r$, such that $b_{i_1} + b_{i_2} + b_{i_3} - (r - l)$ is maximum possible, where $i_1, i_2, i_3$ are the indices of the three maximum elements in range $[l, r]$.

Input

The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 10^5$).

The second line of each test case contains $n$ integers $b_i$ ($1 \leq b_i \leq 10^8$) — beauties of sights $i$ miles from the beginning of the street.

It's guaranteed that the sum of all $n$ does not exceed $10^5$.

Output

For each test case output a single integer equal to the maximum value $b_{i_1} + b_{i_2} + b_{i_3} - (r - l)$ for some running range $[l, r]$.

Example

input

8
1
22
299999996

output

8
1
22
299999996

Note

In the first example, we can choose $l$ and $r$ to be $1$ and $5$. So we visit all the sights and the three sights with the maximum beauty are the sights with indices $1$, $3$, and $5$ with beauties $5$, $4$, and $3$, respectively. So the total value is $5 + 4 + 3 - (5 - 1) = 8$.

In the second example, the range $[l, r]$ can be $[1, 3]$ or $[2, 4]$, the total value is $1 + 1 + 1 - (3 - 1) = 1$.

 

解题思路

　　比赛的时候写了个假双指针的做法，然后动态规划又没想到怎么做。其实只要想到把公式变一下就能做出来了。

　　首先根据贪心思想，对于区间$[l,r]$，若要让$b_{i_1} + b_{i_2} + b_{i_3} - (r - l)$取到最大值，那么其中两个最大元素必然在区间的两个端点，即$a_l, a_r \in \{ {b_{i_1}, b_{i_2}, b_{i_3}} \}$。否则我们可以通过右移左端点$l$或者左移右端点$r$，使得$[l,r]$中的$b_{i_1}, b_{i_2}, b_{i_3}$不变，而$r-l$变小，那么$b_{i_1} + b_{i_2} + b_{i_3} - (r - l)$就会变大。

　　因此，不失一般性，我们假设$i_1 = l$，$i_3 = r$，$i_2 = i \in (l,r)$，那么公式就会变成$b_l + b_i + b_r - (r-l) = a_i + (a_l + l) + (a_r - r)$。可以发现$a_l + l$的取值仅取决于$l$，$a_r - r$的取值仅取决于$r$，因此可以枚举中间元素$a_i$，然后在$l \in [1, i-1]$中求$a_l + l$的最大值，在$r \in [i+1, n]$中求$a_r - r$的最大值。前后缀的最大值可以预处理出来。

　　AC代码如下，时间复杂度为$O(n)$：

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

const int N = 1e5 + 10, INF = 0x3f3f3f3f;

int a[N];
int l[N], r[N];

void solve() {
    int n;
    scanf("%d", &n);
    l[0] = r[n + 1] = -INF;
    for (int i = 1; i <= n; i++) {
        scanf("%d", a + i);
        l[i] = max(l[i - 1], a[i] + i);
    }
    for (int i = n; i; i--) {
        r[i] = max(r[i + 1], a[i] - i);
    }
    int ret = 0;
    for (int i = 1; i <= n; i++) {
        ret = max(ret, a[i] + l[i - 1] + r[i + 1]);
    }
    printf("%d\n", ret);
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        solve();
    }
    
    return 0;
}

 

参考资料

　　Codeforces Round #870 (Div. 2) Editorial：https://codeforces.com/blog/entry/115892