算法1

使用一个set维护当前整个序列中还能够开更号的数的下标

mlogn

时间复杂度

参考文献

C++ 代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<set>
using namespace std;

typedef long long LL;

const int N = 100010;

struct Node{
    int l, r;
    LL sum;
}tr[N * 4];

set<int> pos;

LL w[N];

vector<int> eds;

int n, m;

void pushup(int u)
{
    tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
}

void build(int u, int l, int r)
{
    tr[u] = {l, r};
    if(l == r) 
    {
        tr[u].sum = w[l];
        return;
    }
    int mid = (l + r) / 2;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
    pushup(u);
}

void modify(int u, int x)
{
    if(tr[u].l == tr[u].r)
    {
        tr[u].sum = sqrt(tr[u].sum);
        if(tr[u].sum <= 1) eds.push_back(x);
        return;
    }
    int mid =(tr[u].l + tr[u].r) / 2;
    if(x <= mid) modify(u << 1, x);
    else modify(u << 1 | 1, x);
    pushup(u);
}

LL query(int u, int l, int r)
{
    if(tr[u].l >= l && tr[u].r <= r)
    {
        return tr[u].sum;
    }
    int mid = (tr[u].l + tr[u].r) / 2;
    LL res = 0;
    if(l <= mid) res += query(u << 1, l, r);
    if(r > mid) res += query(u << 1 | 1, l, r);
    return res;
}

int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; ++ i)
    {
        scanf("%lld", &w[i]);
        if(w[i] > 1ll) pos.insert(i);
    }
    build(1, 1, n);

    scanf("%d", &m);
    while(m --)
    {
        int op, l, r;
        scanf("%d %d %d", &op, &l, &r);
        if(op == 1) printf("%lld\n", query(1, l, r));
        else 
        {
            auto iter = pos.lower_bound(l);
            eds.clear();
            while(iter != pos.end() && *iter <= r)
            {
                modify(1, *iter);
                ++ iter;
            }
            for(auto &t: eds) pos.erase(t); 
        }
    }
    return 0;
}