AcWing 245. 你能回答这些问题吗

题目链接：​​传送门​​

线段树单点修改+查询最大子段和
一边写对？
只不过忘了改数组大小没看到而已
维护区间和，区间最大子段和，从左端点开始的最大子段和，从右端点开始的最大子段和
照样维护，取左儿子和右儿子的，还要和左儿子的右儿子的取，对左儿子的 和 左儿子的右儿子的取，对右儿子的 和 右儿子的右儿子的取，这样简单维护起来就好，查询的时候一样

#include <bits/stdc++.h>
#define

using namespace std;
typedef long long ll;
struct node {int l, r, w, f, lf, rf;}tree[A];
int n, m, opt, a, b;
void up(int k) {
  tree[k].w = tree[k << 1].w + tree[k << 1 | 1].w;
  tree[k].f = max(tree[k << 1].f, tree[k << 1 | 1].f);
  tree[k].f = max(tree[k].f, tree[k << 1].rf + tree[k << 1 | 1].lf);
  tree[k].lf = max(tree[k << 1].lf, tree[k << 1].w + tree[k << 1 | 1].lf);
  tree[k].rf = max(tree[k << 1 | 1].rf, tree[k << 1 | 1].w + tree[k << 1].rf);
}
void build(int k, int l, int r) {
  tree[k].l = l; tree[k].r = r;
  if (l == r) {
    int d; scanf("%d", &d);
    tree[k].w = tree[k].f = tree[k].lf = tree[k].rf = d;
    return;
  }
  int m = (l + r) >> 1;
  build(k << 1, l, m); build(k << 1 | 1, m + 1, r);
  up(k);
}
void change(int k, int pos, int val) {
  if (tree[k].l == tree[k].r) {
    tree[k].w = tree[k].f = tree[k].lf = tree[k].rf = val;
    return;
  }
  int m = (tree[k].l + tree[k].r) >> 1;
  if (pos <= m) change(k << 1, pos, val);
  else change(k << 1 | 1, pos, val);
  up(k);
}
node ask(int k, int l, int r) {
  if (tree[k].l >= l and tree[k].r <= r) return tree[k];
  int m = (tree[k].l + tree[k].r) >> 1;
  if (l <= m and r > m) {
    node lx = ask(k << 1, l, r), rx = ask(k << 1 | 1, l, r), tmp;
    tmp.f = max(lx.f, rx.f); tmp.f = max(tmp.f, lx.rf + rx.lf);
    tmp.lf = max(lx.lf, lx.w + rx.lf); tmp.rf = max(rx.rf, rx.w + lx.rf);
    return tmp;
  }
  if (l <= m) return ask(k << 1, l, r);
  if (r > m) return ask(k << 1 | 1, l, r);
}

int main(int argc, char const *argv[]) {
  cin >> n >> m; build(1, 1, n);
  while (m--) {
    scanf("%d%d%d", &opt, &a, &b);
    if (opt == 1) printf("%d\n", ask(1, min(a, b), max(a, b)).f);
    else change(1, a, b);
  }
  return 0;
}