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>
typedef long long ll;
template <typename T> class Vector : public std::vector<T> {
public:
using std::vector<T>::vector;
std::vector<ll> psum;
void calculate_prefix_sum() {
psum.resize(this->size());
std::partial_sum(this->begin(), this->end(), psum.begin());
}
T sum(ll a, ll b) {
if (a > b) {
return 0;
}
T ans = psum[b];
if (a - 1 >= 0) {
ans -= psum[a - 1];
}
return ans;
}
};
const ll N = 1e5 + 1, K = 201;
const ll inf = 1e15;
ll dp[K][N];
struct Line {
ll m, b;
ll eval(ll x) {
return m * x + b;
}
};
class CHT {
private:
bool parallel(const Line &a, const Line &b) {
return a.m == b.m;
}
long double x_intersection(const Line &a, const Line &b) {
return (a.b - b.b) / ((long double) (b.m - a.m));
}
public:
std::deque<Line> dq;
void insert(Line l) {
if (dq.size() >= 1 && parallel(l, dq.front()) && l.b <= dq.front().b) {
return;
}
while (dq.size() >= 2) {
if (parallel(l, dq.front())) {
if (l.b > dq.front().b) {
dq.pop_front();
continue;
} else {
return;
}
}
if (x_intersection(l, dq[0]) >= x_intersection(dq[0], dq[1])) {
dq.pop_back();
} else {
break;
}
}
if (dq.size() == 1 && parallel(l, dq.front()) && l.b > dq.front().b) {
dq.pop_front();
}
dq.push_front(l);
}
ll query(ll x) {
while (dq.size() >= 2 && dq[0].eval(x) < dq[1].eval(x)) {
dq.pop_front();
}
return dq[0].eval(x);
}
};
int main() {
ll n, k;
std::cin >> n >> k;
Vector<ll> a(n);
for (auto &i : a) {
std::cin >> i;
}
a.calculate_prefix_sum();
if (n == 2) {
std::cout << a[0] * a[1] << "\n1\n";
return 0;
}
for (ll i = 0; i <= k; ++ i) {
CHT hull;
for (ll j = n - 1; j >= 0; -- j) {
if (j == n - 1) {
if (i != 0) {
dp[i][j] = -inf;
} else {
dp[i][j] = 0;
}
continue;
}
if (i == 0) {
dp[i][j] = 0;
continue;
}
ll m = a.sum(j + 1, n - 1);
ll b = dp[i - 1][j + 1] + a.sum(0, j) * a.sum(j + 1, n - 1);
// slope is monotonically increasing
hull.insert({m, b});
// queries are monotonically increasing
dp[i][j] = std::max(0LL, hull.query(-a.sum(0, j - 1)));
}
}
std::cout << dp[k][0] << "\n";
ll i = 0;
ll ops = k;
for (ll c = i; c < n - 1 && ops > 0; c++) {
if (dp[ops][i] == a.sum(i, c) * a.sum(c + 1, n - 1) + dp[ops - 1][c + 1]) {
std::cout << c + 1 << " ";
ops--;
i = c + 1;
}
}
}
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