D. Yet Another Yet Another Task (ST表模版 + 单调队列)

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <ctime>
#include <bitset>
#include <cmath>
#include <sstream>
#include <iostream>

#define ll long long
#define ls nod<<1
#define rs (nod<<1)+1
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define INF 0x3f3f3f3f3f3f3f3f
#define max(a, b) (a>b?a:b)
#define min(a, b) (a<b?a:b)


const double eps = 1e-10;
const int maxn = 2e5 + 10;
const int MOD = 998244353;

int sgn(double a) { return a < -eps ? -1 : a < eps ? 0 : 1; }

using namespace std;


int a[maxn];
int q[maxn];
int f1[maxn],f2[maxn];


int pre[maxn],maxx[maxn][20],minn[maxn][20];
void rmq(int n)
{
    for(int i=0;i<=n;i++) minn[i][0]=maxx[i][0]=pre[i];
    for(int j=1;(1<<j)<=n;j++)
    {
        for(int i=0;i+j-1<=n;i++)
        {
            maxx[i][j]=max(maxx[i][j-1],maxx[i+(1<<(j-1))][j-1]);
            minn[i][j]=min(minn[i][j-1],minn[i+(1<<(j-1))][j-1]);
        }
    }
}
int q1(int l,int r)
{
    int k=0,dis=r-l+1;
    while((1<<(k+1))<=dis) ++k;
    return max(maxx[l][k],maxx[r-(1<<k)+1][k]);
}
int q2(int l,int r)
{
    int k=0,dis=r-l+1;
    while((1<<(k+1))<=dis) ++k;
    return min(minn[l][k],minn[r-(1<<k)+1][k]);
}


int main() {
    int n;
    scanf("%d",&n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        pre[i] = pre[i-1] + a[i];
    }
    rmq(n);
    int cnt;

    // 右边第一个 > a[i] 的数的位置
    q[1] = n + 1;
    cnt = 1;
    a[n + 1] = INF;
    for (int i = n; i >= 1; i--) {
        while (cnt && a[q[cnt]] <= a[i])
            cnt--;
        f1[i] = q[cnt] - 1;
        q[++cnt] = i;
    }
    // 左边第一个 > a[i] 的数的位置
    q[1] = 0;
    cnt = 1;
    a[0] = INF;
    for (int i = 1;i <= n;i++) {
        while (cnt && a[q[cnt]] <= a[i])
            cnt--;
        f2[i] = q[cnt];
        q[++cnt] = i;
    }
//    for (int i = 1;i <= n;i++) {
//        cout << f2[i] << " ";
//    }
//    cout << endl;
    int ans = 0;
    for (int i = 1;i <= n;i++) {
        int l = f2[i],r = f1[i];
        int maxl = q2(l,i-1);
        int maxr = q1(i,r);
        ans = max(ans,maxr - maxl - a[i]);
    }
    printf("%d\n",ans);
    return 0;
}