This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
class FenwickTree {
public:
std::vector<long long> f;
int n;
FenwickTree(int n) : n(n) { f.resize(n + 1); }
void add(int idx, int del) {
for (int i = idx; i <= n; i += i & (-i)) {
f[i] += del;
}
}
long long query(int idx) {
long long ans = 0;
for (int i = idx; i >= 1; i -= i & (-i)) {
ans += f[i];
}
return ans;
}
};
int main() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(nullptr);
int n, m;
std::cin >> n >> m;
std::vector owned(n + 1, std::vector<int>());
for (int i = 1, x; i <= m; ++i) {
std::cin >> x;
owned[x].push_back(i);
}
std::vector<long long> p(n + 1);
for (int i = 1; i <= n; ++i) {
std::cin >> p[i];
}
int k;
std::cin >> k;
struct Update {
int l, r, a;
};
std::vector<Update> upd(k);
for (auto &[l, r, a] : upd) {
std::cin >> l >> r >> a;
}
struct State {
int i, l, r;
};
std::vector midpoints(k, std::vector<State>());
for (int i = 0; i < n; ++i) {
midpoints[(k - 1) / 2].push_back({i, 0, k - 1});
}
std::vector<int> ans(n, k);
FenwickTree tree(m + 1);
auto apply = [&](int l, int r, int d) {
tree.add(l, d);
tree.add(r + 1, -d);
};
while (true) {
bool changed = false;
for (int i = 0; i < k; ++i) {
if (upd[i].l <= upd[i].r) {
apply(upd[i].l, upd[i].r, upd[i].a);
} else {
apply(upd[i].l, m, upd[i].a);
apply(1, upd[i].r, upd[i].a);
}
while (!midpoints[i].empty()) {
changed = true;
auto [_i, l, r] = midpoints[i].back();
midpoints[i].pop_back();
if (l > r) {
continue;
}
long long obtained = 0;
for (auto &x : owned[_i + 1]) {
obtained += tree.query(x);
if (obtained >= p[_i + 1]) {
break;
}
}
if (obtained >= p[_i + 1]) {
ans[_i] = std::min(ans[_i], i);
}
if (l < r) {
if (obtained >= p[_i + 1] and l + i - 1 >= 0) {
midpoints[(l + i - 1) / 2].push_back({_i, l, i - 1});
}
if (obtained < p[_i + 1] and i + r + 1 < 2 * k) {
midpoints[(i + r + 1) / 2].push_back({_i, i + 1, r});
}
}
}
}
if (!changed) {
break;
}
tree.f.clear();
}
for (auto &i : ans) {
if (i == k) {
std::cout << "NIE\n";
} else {
std::cout << i + 1 << '\n';
}
}
}
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