// Success consists of going from failure to failure without loss of enthusiasm
#include <bits/stdc++.h>
using namespace std;
#define nl '\n'
#define pb push_back
#define mp make_pair
#define f first
#define s second
#define sz(x) int(x.size())
template<class T> using V = vector<T>;
using vi = V<int>;
using ll = long long;
using pl = pair<ll, ll>;
using vl = V<ll>;
// using vpi = V<pi>;
const ll INFL = ll(1e16) + 1008;
const int nax = 1e6 + 6;
const int SQ = 1e3 + 1;
pl oc[nax];
int main() {
cin.tie(0)->sync_with_stdio(0);
for(int x = 0; x < nax; x++) {
oc[x] = mp(INFL, -1);
}
int N; cin >> N;
vi A(N); for(auto& x : A) cin >> x;
vi W(N); for(auto& x : W) cin >> x;
vl P = {0}; for(auto& x : W) P.pb(P.back() + x);
auto query = [&](int l, int r) {
if (l > r) return 0LL;
return P[r + 1] - P[l];
};
auto sq = [&](int x) { return x * 1LL * x; };
vl dp(N, INFL); dp[0] = 0;
oc[A[0]] = mp(dp[0] - P[0], 0);
for(int i = 1; i < N; i++) {
int x = A[i];
for(int d = max(-SQ, -x); d <= min(SQ, nax-1-x); d++) {
int y = x + d;
if (oc[y].second == -1) continue;
int j = oc[y].second; ll cost = sq(x - y) + query(j + 1, i - 1);
cout << i << " " << j << " => " << cost << endl;
dp[i] = min(dp[i], dp[j] + cost);
}
ll c = dp[i] - P[i];
if (oc[x].first > c) oc[x] = mp(c, i);
}
cout << dp.back() << nl;
return 0;
}
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Incorrect |
7 ms |
15956 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Execution timed out |
3053 ms |
70024 KB |
Time limit exceeded |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Incorrect |
7 ms |
15956 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |