이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define ST first
#define ND second
#define PB push_back
using namespace std;
using ll = long long;
using pi = pair<int, int>;
using vi = vector<int>;
const int nax = 300 * 1000 + 10;
struct SegTree {
ll T[(1 << 20)];
int R;
void upd(int a, ll x) {
a += R;
T[a] += x;
while (a > 1) {
a >>= 1;
T[a] += x;
}
}
ll qr(int a, int b) {
a += R; b += R;
if (a > b) return 0LL;
ll w = T[a] + (a != b ? T[b] : 0LL);
while (a/2 != b/2) {
if (a % 2 == 0) w += T[a + 1];
if (b % 2 == 1) w += T[b - 1];
a >>= 1;
b >>= 1;
}
return w;
}
void init(int n) {
R = 1;
while (R < n) R <<= 1;
for (int i = 0; i < 2 * R; ++i) T[i] = 0;
}
int go_down() {
int v = 1;
while (v < R) {
if (T[2 * v + 1] != 0) v = v*2 + 1;
else v = v*2;
}
return v - R;
}
};
vector<pair<int,int>>qrs[nax];
ll ans[nax];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int n, q;
ll d;
cin >> n >> d;
vector<ll> a(n);
for (ll &x : a) cin >> x;
cin >> q;
vector<ll> demands(n);
for (int i = 1; i < n; ++i) {
demands[i] = a[i] - a[i - 1];
if (demands[i] < 0) {
demands[i] = (abs(demands[i]) - 1) / d + 1;
} else {
demands[i] = -(demands[i] / d);
}
}
for (int l,r, i = 0; i < q; ++i) {
cin >> l >> r;
l--, r--;
qrs[r].emplace_back(l, i);
}
SegTree s, lft, closed, total;
s.init(n);
lft.init(n);
closed.init(n);
total.init(n);
for (int i = 1; i < n; ++i) {
// update value i
//cerr << "i: " << i << " " << demands[i] << "\n";
if (demands[i] < 0) {
s.upd(i, -demands[i]); // I can give that many
} else {
lft.upd(i, demands[i] * i);
total.upd(i, demands[i]);
while (demands[i] > 0) {
if (s.T[1] == 0) break;
int v = s.go_down();
ll me = min(demands[i], s.qr(v,v));
s.upd(v, -me);
closed.upd(v, me);
lft.upd(v, -me * v);
//cerr << "segment: " << v << "->" << i << " " << me << " times\n";
}
}
for (auto [l, id] : qrs[i]) {
ll open = total.qr(l + 1, i) - closed.qr(l + 1, i);
ll res = lft.qr(l + 1, i);
ll me = a[l] / d;
//cerr << "(" << total.qr(l+1,i) << " " << closed.qr(l+1, i) << ")" << " " << res << " " << me << "\n";
if (me < open) {
ans[id] = -1;
} else {
ans[id] = res - (ll)l * open;
}
}
}
for (int i = 0; i < q; ++i) cout << ans[i] << "\n";
}
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