CF438D The Child and Sequence

题目链接：​​传送门​​

区间求和，单点修改，区间取模

线段树里维护一个max，取模的时候如果区间最大值小于max直接return就可以
不然的话递归到叶子节点再取模
有效降低复杂度
不开long long好像只能过两个样例

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <complex>
#include <algorithm>
#include <climits>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define
#define

using namespace std;
typedef long long ll;
#define
struct node {
  int l, r, w, mx;
}tree[A];
int n, m, a, b, c, aa[A];
void up(int k) {
  tree[k].w = tree[k << 1].w + tree[k << 1 | 1].w;
  tree[k].mx = max(tree[k << 1].mx, tree[k << 1 | 1].mx);
}
void build(int k, int l, int r) {
  tree[k].l = l; tree[k].r = r;
  if (l == r) {
    tree[k].w = tree[k].mx = aa[l];
    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].mx = 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);
}
int ask(int k, int l, int r) {
  if (tree[k].l >= l and tree[k].r <= r) return tree[k].w;
  int m = (tree[k].l + tree[k].r) >> 1, ans = 0;
  if (l <= m) ans += ask(k << 1, l, r);
  if (r > m) ans += ask(k << 1 | 1, l, r);
  return ans;
}
void ccc(int k, int l, int r, int mod) {
  if (tree[k].mx < mod) return;
  if (tree[k].l == tree[k].r) {
    tree[k].w %= mod;
    tree[k].mx %= mod;
    return;
  }
  int m = (tree[k].l + tree[k].r) >> 1;
  if (l <= m) ccc(k << 1, l, r, mod);
  if (r > m) ccc(k << 1 | 1, l, r, mod);
  up(k);
}

signed main(signed argc, char const *argv[]) {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> aa[i];
  build(1, 1, n); int opt;
  while (m--) {
    cin >> opt;
    if (opt == 1) {
      cin >> a >> b;
      cout << ask(1, a, b) << endl;
    }
    if (opt == 2) {
      cin >> a >> b >> c;
      ccc(1, a, b, c);
    }
    if (opt == 3) {
      cin >> a >> b;
      change(1, a, b);
    }
  }
  return 0;
}