1 条题解
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0
线段树
注意开 long long
#include <bits/stdc++.h> #define lowbit(x) (x & (-x)) typedef long long ll; using namespace std; const int N = 5e5 + 1; ll n, m, a[N]; ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; } struct Node { int l, r; ll sum, d; } tr[N << 2]; void pushup(Node& u, Node& l, Node& r) { u.sum = l.sum + r.sum; u.d = gcd(l.d, r.d); } void pushup(int u) { pushup(tr[u], tr[u << 1], tr[u << 1 | 1]); } void build(int u, int l, int r) { if (l == r) { ll b = a[l] - a[l - 1]; tr[u] = {l, r, b, b}; return; } tr[u] = {l, r}; int mid = l + r >> 1; build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r); pushup(u); } void modify(int u, int x, ll v) { if (tr[u].l == x && tr[u].r == x) { ll b = tr[u].sum + v; tr[u] = {x, x, b, b}; return; } int mid = tr[u].l + tr[u].r >> 1; if (x <= mid) modify(u << 1, x, v); else modify(u << 1 | 1, x, v); pushup(u); } Node query(int u, int l, int r) { if (tr[u].l >= l && tr[u].r <= r) return tr[u]; int mid = tr[u].l + tr[u].r >> 1; if (mid >= r) return query(u << 1, l, r); else if (mid < l) return query(u << 1 | 1, l, r); else { Node left = query(u << 1, l, r), right = query(u << 1 | 1, l, r); Node res; pushup(res, left, right); return res; } } int main() { scanf("%lld%lld", &n, &m); for (int i = 1; i <= n; i++) scanf("%lld", &a[i]); build(1, 1, n); int l, r; ll d; char op[2]; while (m--) { scanf("%s%d%d", op, &l, &r); if (*op == 'C') { scanf("%lld", &d), modify(1, l, d); if (r + 1 <= n) modify(1, r + 1, -d); } else { Node left = query(1, 1, l); Node right = {0, 0, 0ll, 0ll}; if (l + 1 <= r) right = query(1, l + 1, r); printf("%lld\n", abs(gcd(left.sum, right.d))); } } return 0; }
- 1
信息
- ID
- 1547
- 时间
- 1000ms
- 内存
- 256MiB
- 难度
- 10
- 标签
- 递交数
- 16
- 已通过
- 1
- 上传者