BZOJ#3038上帝造题的七分钟2

上帝造题的七分钟2

思路：

数据范围\(10^{12}\),大概开方5六次之后就变成1，当一个区间全是1的时候就没必要改变，其他暴力修改即可

代码：

#include <bits/stdc++.h>
#define int long long
int _ = 0, Case = 1;
using namespace std;
#define all(v) begin(v),end(v)
#define nline '\n'
const int N = 100010;
int a[N];
struct T {
    int l, r, val, tag;
} tr[N << 2];
void pushup(int p) {
    tr[p].val = tr[p << 1].val + tr[p << 1 | 1].val;
    tr[p].tag = tr[p << 1].tag + tr[p << 1 | 1].tag;
}
void build(int p, int l, int r) {
    if (l == r) {
        tr[p] = {l, r, a[l], 0};
        return;
    }
    tr[p] = {l, r, 0, 0};
    int mid = l + r >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    pushup(p);
}
int query(int p, int l, int r) {
    if (tr[p].l >= l and tr[p].r <= r) return tr[p].val;
    int mid = tr[p].l + tr[p].r >> 1;
    if (l > mid) return query(p << 1 | 1, l, r);
    else if (r <= mid) return query(p << 1, l, r);
    return query(p << 1, l, r) + query(p << 1 | 1, l, r);
}
void modify(int p, int l, int r) {
    if (tr[p].tag == tr[p].r - tr[p].l + 1) {
        return;
    }
    else {
        if (tr[p].l == tr[p].r) {
            if (tr[p].val <= 1) {
                tr[p].tag = 1;
            } else {
                tr[p].tag = 0;
            }
            tr[p].val = sqrt(tr[p].val);
            if (tr[p].val <= 1) tr[p].tag = 1;
            else tr[p].tag = 0;
            return;
        } else {
            int mid = tr[p].l + tr[p].r >> 1;
            if (l <= mid) modify(p<<1, l, r);
            if (r > mid) modify(p<<1|1, l, r);
            pushup(p);
        }
    }
}
void solve(int Case) {
    int n, m;
    cin >> n ;
    for (int i = 1; i <= n; i++) cin >> a[i];
    build(1, 1, n);
    cin >> m;
    for (; m--;) {
        int k, l, r;
        cin >> k >> l >> r;
        if (!k) modify(1, min(l, r), max(l, r));
        else cout << query(1, min(l, r), max(l, r)) << nline;

    }
}

signed main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
//   cin >> _; for (Case = 1; Case <= _; Case++)
    solve(Case);

    return 0;
}