Problem:
Score : 700 points

Problem Statement

We have a rooted binary tree with N vertices, where the vertices are numbered 1 to N.
Vertex 1 is the root, and the parent of Vertex i (i \geq 2) is Vertex \left[ \frac{i}{2} \right].

Each vertex has one item in it. The item in Vertex i has a value of V_i and a weight of W_i.
Now, process the following query Q times:

Given are a vertex v of the tree and a positive integer L.
Let us choose some (possibly none) of the items in v and the ancestors of v so that their total weight is at most L.
Find the maximum possible total value of the chosen items.

Here, Vertex u is said to be an ancestor of Vertex v when u is an indirect parent of v, that is, there exists a sequence of vertices w_1,w_2,\ldots,w_k (k\geq 2) where w_1=v, w_k=u, and w_{i+1} is the parent of w_i for each i.

Constraints

All values in input are integers.

1 \leq N < 2^{18}

1 \leq Q \leq 10^5

1 \leq V_i \leq 10^5

1 \leq W_i \leq 10^5

For the values v and L given in each query, 1 \leq v \leq N and 1 \leq L \leq 10^5.

Input

Let v_i and L_i be the values v and L given in the i-th query.
Then, Input is given from Standard Input in the following format:

N
V_1 W_1
:
V_N W_N
Q
v_1 L_1
:
v_Q L_Q

Output

For each integer i from 1 through Q,
the i-th line should contain the response to the i-th query.

Sample Input 1

3
1 2
2 3
3 4
3
1 1
2 5
3 5

Sample Output 1

0
3
3

In the first query, we are given only one choice: the item with (V, W)=(1,2). Since L = 1, we cannot actually choose it, so our response should be 0.

In the second query, we are given two choices: the items with (V, W)=(1,2) and (V, W)=(2,3). Since L = 5, we can choose both of them, so our response should be 3.

Sample Input 2

15
123 119
129 120
132 112
126 109
118 103
115 109
102 100
130 120
105 105
132 115
104 102
107 107
127 116
121 104
121 115
8
8 234
9 244
10 226
11 227
12 240
13 237
14 206
15 227

Sample Output 2

256
255
250
247
255
259
223
253

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)


#include <bits/stdc++.h>
//#include <boost/multiprecision/cpp_int.hpp>
//namespace mp = boost::multiprecision;

using namespace std;

const double PI = 3.14159265358979323846;
typedef long long ll;
const double EPS = 1e-9;
#define rep(i, n) for (int i = 0; i < (n); ++i)
//#define rep(i, n) for (ll i = 0; i < (n); ++i)
typedef pair<ll, ll> P;
const ll INF = 10e17;
#define cmin(x, y) x = min(x, y)
#define cmax(x, y) x = max(x, y)
#define ret() return 0;

double equal(double a, double b) {
    return fabs(a - b) < DBL_EPSILON;
}

std::istream &operator>>(std::istream &in, set<int> &o) {
    int a;
    in >> a;
    o.insert(a);
    return in;
}

std::istream &operator>>(std::istream &in, queue<int> &o) {
    ll a;
    in >> a;
    o.push(a);
    return in;
}

bool contain(set<int> &s, int a) { return s.find(a) != s.end(); }

//ofstream outfile("log.txt");
//outfile << setw(6) << setfill('0') << prefecture << setw(6) << setfill('0') << rank << endl;
// std::cout << std::bitset<8>(9);
//const ll mod = 1e10;

typedef priority_queue<ll, vector<ll>, greater<ll> > PQ_ASK;

struct Item {
    ll v, w;
};


std::istream &operator>>(std::istream &in, Item &o) {
    cin >> o.v >> o.w;
    return in;
}

vector<int> create_candidates(int v) {
    vector<int> u;
    u.push_back(v);
    while (v > 0) {
        v = (v + 1) / 2 - 1;
        u.push_back(v);
    }
    return u;
}

vector<P> create_table(vector<Item> &use_items) {
    unordered_map<ll, ll> m;
    int n = use_items.size();
    rep(i, 1 << n) {
        ll w = 0, v = 0;
        rep(j, n) {
            if (!((i >> j) & 1)) continue;
            w += use_items[j].w;
            v += use_items[j].v;
        }
        cmax(m[w], v);
    }

    vector<P> ans;
    ans.push_back({0, 0});

    vector<ll> keys;
    for (auto &e : m) keys.push_back(e.first);
    sort(keys.begin(), keys.end());

    for (ll key : keys) {
        if (key == 0) continue;
        ll value = m[key];
        if (ans.back().second >= value) continue;
        ans.push_back({key, value});
    }

    return ans;
}

void from_cache(ll u, ll l, unordered_map<ll, vector<P>> &cache, vector<Item> &items) {
    auto index = create_candidates(u);

    auto it = upper_bound(cache[u].begin(), cache[u].end(), P(l, INF));
    it--;
    ll ans = it->second;
    cout << ans << endl;

}

void use_cache(ll u, ll l, unordered_map<ll, vector<P>> &cache, int fe, vector<Item> &items) {
    auto index = create_candidates(u);

    int fm = [&] {
        int ans = 0;
        for (int i : index) if (i <= fe) cmax(ans, i);
        return ans;
    }();

    vector<Item> overs;
    for (int i : index) if (i > fe) overs.push_back(items[i]);

    auto table1 = create_table(overs);

    ll ans = 0;
    for (auto e1 : table1) {
        if (e1.first > l) continue;
        ll sub = l - e1.first;
        auto it = upper_bound(cache[fm].begin(), cache[fm].end(), P(sub, INF));

        it--;
        assert(it->first + e1.first <= l);
        ll now = e1.second + it->second;
        cmax(ans, now);
    }
    cout << ans << endl;

}

int main() {


    unordered_map<ll, vector<P>> cache;

    int n;
    cin >> n;
    vector<Item> items(n);
    rep(i, n) cin >> items[i];


    int f = 8;//log2(n) / 2;
    int fe = (1 << (f + 1)) - 2;

    for (int i = 0; i <= fe && i < n; i++) {
        auto index = create_candidates(i);
        vector<Item> use_items;
        for (int j : index) use_items.push_back(items[j]);
        cache[i] = create_table(use_items);
    }

    int q;
    cin >> q;
    rep(_, q) {
        int u, l;
        cin >> u >> l;
        u--;

        if (u <= fe) {
            from_cache(u, l, cache, items);
        } else {
            use_cache(u, l, cache, fe, items);
        }
    }

    cout << endl;
}