牛客-小阳的贝壳

题目传送门

sol：题目读完就知道是线段树，gcd满足结合律。操作2也好想到差分，但是不会修改后维护gcd。看了题解发现还是差分，这个差分用的妙啊；

• 线段树+差分

  #include "bits/stdc++.h"
  using namespace std;
  const int MAXN = 1e5 + 10;
  struct Tree{
      int l, r;
      int sum;
      int sub;
      int gcd;
  } segTree[MAXN * 4];
  int arr[MAXN];
  void build(int i, int l, int r) {
      segTree[i].l = l;
      segTree[i].r = r;
      if (l == r) return;
      int mid = l + r >> 1;
      build(i << 1, l, mid);
      build(i << 1 | 1, mid + 1, r);
  }
  void update(int i, int index, int val) {
      if (segTree[i].l == segTree[i].r) {
          segTree[i].sum = val;
          segTree[i].sub = max(val, -val);
          segTree[i].gcd = segTree[i].sub;
          return;
      }
      int mid = segTree[i].l + segTree[i].r >> 1;
      if (index <= mid) update(i << 1, index, val);
      else update(i << 1 | 1, index, val);
      segTree[i].sum = segTree[i << 1].sum + segTree[i << 1 | 1].sum;
      segTree[i].sub = max(segTree[i << 1].sub, segTree[i << 1 | 1].sub);
      segTree[i].gcd = __gcd(segTree[i << 1].gcd, segTree[i << 1 | 1].gcd);
  }
  int query(int i, int op, int l, int r) {
      if (l <= segTree[i].l && r >= segTree[i].r) {
          if (op == 1) return segTree[i].sum;
          if (op == 2) return segTree[i].sub;
          if (op == 3) return segTree[i].gcd;
      }
      int mid = segTree[i].l + segTree[i].r >> 1;
      int ans = 0;
      if (op == 1) {
          if (l <= mid) ans += query(i << 1, 1, l, r);
          if (r > mid) ans += query(i << 1 | 1, 1, l, r);
      } else if (op == 2) {
          if (l <= mid) ans = max(ans, query(i << 1, 2, l, r));
          if (r > mid) ans = max(ans, query(i << 1 | 1, 2, l, r));
      } else {
          if (l <= mid) ans = __gcd(ans, query(i << 1, 3, l, r));
          if (r > mid) ans = __gcd(ans, query(i << 1 | 1, 3, l, r));
      }
      return ans;
  }
  int main() {
      int n, m; 
      scanf("%d%d", &n, &m); build(1, 1, n);
      for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
      for (int i = n; i >= 1; i--) {
          arr[i] -= arr[i - 1];
          update(1, i, arr[i]);
      }
      while (m--) {
          int op, l, r, x;
          scanf("%d", &op);
          if (op == 1) {
              scanf("%d%d%d", &l, &r, &x);
              arr[l] += x;
              update(1, l, arr[l]);
              if (r + 1 <= n) {
                  arr[r + 1] -= x;
                  update(1, r + 1, arr[r + 1]);
              }
          } else if (op == 2) {
              scanf("%d%d", &l, &r);
              if (l == r) puts("0");
              else printf("%d
  ", query(1, 2, l + 1, r));
          } else {
              scanf("%d%d", &l, &r);
              if (l == r) printf("%d
  ", query(1, 1, 1, l));
              else printf("%d
  ", __gcd(query(1, 1, 1, l), query(1, 3, l + 1, r)));
          }
      }
      return 0;
  }