이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <cmath>
#include <limits>
#include <queue>
#include <string.h>
#include <stack>
#include <limits>
using namespace std;
int N, K;
long long p[100005], dp[201][100005];
long long minf = numeric_limits<long long>::min();
struct seg {
long long k = 0, b = minf;
seg() {}
seg(long long _k, long long _b) :k(_k), b(_b) {}
long long f(long long x) {
//cout << "k : " << k << " x : " << x << " b : " << b << endl;
return k * x + b;
}
};
struct Node {
Node * left = nullptr;
Node * right = nullptr;
seg s;
~Node() {
if(left != nullptr) delete left;
if(right != nullptr) delete right;
}
};
// max value
void add(Node* cur, seg ns, int nl, int nr) {
if(cur->s.b == minf && ns.b == minf) return;
int m = (nl + nr) / 2;
bool lef = cur->s.f(nl) < ns.f(nl);
bool mid = cur->s.f(m) < ns.f(m);
//cout << "cur : " << cur->s.f(m) << " ns : " << ns.f(m) << " result : " << (cur->s.f(m) < ns.f(m)) << endl;
if(mid) swap(cur->s, ns);
if(nl == nr) return;
else if(lef != mid) {
if(cur->left == nullptr) cur->left = new Node();
add(cur->left, ns, nl, m);
}
else {
if(cur->right == nullptr) cur->right = new Node();
add(cur->right, ns, m + 1, nr);
}
}
long long query(Node* cur, int x, int nl, int nr) {
if(nl == nr) {return cur->s.f(x);}
int m = (nl + nr) / 2;
if(x <= m) {
if(cur->left == nullptr) {return cur->s.f(x);}
return max(cur->s.f(x), query(cur->left, x, nl, m));
} else {
if(cur->right == nullptr) {return cur->s.f(x);}
return max(cur->s.f(x), query(cur->right, x, m + 1, nr));
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> N >> K;
long long q;
for(int i = 1; i <= N; ++i) {
cin >> q;
p[i] = p[i - 1] + q;
}
Node * tree;
long long ans = 0;
for(int k = 1; k <= K; ++k) {
tree = new Node();
for(int i = N - k - 1; i >= 0; --i){
add(tree, seg(p[i + 1], - p[i + 1] * p[i + 1] + p[N] * p[i + 1] + dp[k - 1][i + 1]), 0, 1e9);
dp[k][i] = query(tree, p[i], 0, 1000 * N) - p[N] * p[i];
if(k == K) ans = max(ans, dp[k][i]);
}
delete(tree);
}
cout << ans << endl;
int s = 0;
for(int k = K; k >= 1; --k) {
for(int i = s + 1; i <= N - k; ++i) {
if(ans - ( (p[N] - p[i]) * (p[i] - p[s]) ) == dp[k - 1][i] ) {
cout << i << " ";
ans -= (p[N] - p[i]) * (p[i] - p[s]);
s = i;
break;
}
}
}
return 0;
}
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