Can you answer these queries III

题意

$n$个数，$q$ 次操作

操作$0xy$把$A_x$修改为$y$

操作$1lr$询问区间$[l,r]$的最大子段和

输入输出样例

输入

4
1 2 3 4
4
1 1 3
0 3 -3
1 2 4
1 3 3

输出

6
4
-3

/***  Amber  ***/
//#pragma GCC optimize(3,"Ofast","inline")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define ls (rt<<1)
#define rs (rt<<1|1)
const double Pi=acos(-1);
const double eps=1e-6;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
const int maxn = 5e4+10;

int n;
struct node{
     
    int sum,lsum,rsum,ans;
}t[maxn << 2];
int a[maxn];
void pushup(int rt) {
     
    t[rt].sum = t[ls].sum + t[rs].sum;
    t[rt].lsum = max(t[ls].sum + t[rs].lsum, t[ls].lsum);
    t[rt].rsum = max(t[rs].sum + t[ls].rsum, t[rs].rsum);
    t[rt].ans = max(t[ls].ans, max(t[rs].ans, t[ls].rsum + t[rs].lsum));
}
void build(int rt,int l,int r) {
     
    if (l == r) {
     
        t[rt].ans = t[rt].lsum = t[rt].rsum = t[rt].sum = a[l];
        return;
    }
    int mid = l + r >> 1;
    build(ls, l, mid);
    build(rs, mid + 1, r);
    pushup(rt);
}
void change(int rt,int l,int r,int x,int k) {
     
    if (l == r) {
     
        t[rt].ans = t[rt].lsum = t[rt].rsum = t[rt].sum = k;
        return;
    }
    int mid = l + r >> 1;
    if (x <= mid) change(ls, l, mid, x, k);
    else change(rs, mid + 1, r, x, k);
    pushup(rt);
}
node query(int rt,int l,int r,int ql,int qr) {
     
    if (ql <= l && r <= qr) {
     
        return t[rt];
    }
    int mid = l + r >> 1;
    if (qr <= mid) return query(ls, l, mid, ql, qr);
    if (ql > mid) return query(rs, mid + 1, r, ql, qr);
    node L = query(ls, l, mid, ql, qr), R = query(rs, mid + 1, r, ql, qr);

    node res;
    res.sum = L.sum + R.sum;
    res.ans = max(L.ans,max(R.ans,R.lsum+L.rsum));
    res.lsum = max(L.lsum, L.sum + R.lsum);
    res.rsum = max(R.rsum, R.sum + L.rsum);
    return res;
}
inline void work() {
     
    scanf("%d",&n);
    for (int i = 1; i <= n; i++) {
     
        scanf("%d",&a[i]);
    }
    build(1, 1, n);
    int q;
    scanf("%d",&q);
    while (q--) {
     
        int op, x, y;
        scanf("%d%d%d",&op,&x,&y);
        if (op == 0) {
     
            change(1, 1, n, x, y);
        } else {
     
            printf("%d\n", query(1, 1, n, x, y).ans);
        }
    }
}
int main() {
     
    work();
    return 0;
}