1 条题解

  • 0
    @ 2024-10-7 11:20:50

    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