제출 #1301533

#제출 시각아이디문제언어결과실행 시간메모리
1301533Nonoze수열 (APIO14_sequence)C++20
100 / 100
611 ms82692 KiB
/* * Author: Nonoze * Created: Tuesday 18/11/2025 */ #include <bits/stdc++.h> using namespace std; #ifndef DEBUG #define dbg(...) #endif // #define cout cerr << "OUT: " #define endl '\n' #define endlfl '\n' << flush #define quit(x) return (void)(cout << x << endl) template<typename T> void read(T& x) { cin >> x; } template<typename T1, typename T2> void read(pair<T1, T2>& p) { read(p.first), read(p.second); } template<typename T> void read(vector<T>& v) { for (auto& x : v) read(x); } template<typename T1, typename T2> void read(T1& x, T2& y) { read(x), read(y); } template<typename T1, typename T2, typename T3> void read(T1& x, T2& y, T3& z) { read(x), read(y), read(z); } template<typename T1, typename T2, typename T3, typename T4> void read(T1& x, T2& y, T3& z, T4& zz) { read(x), read(y), read(z), read(zz); } template<typename T> void print(vector<T>& v) { for (auto& x : v) cout << x << ' '; cout << endl; } #define sz(x) (int)(x.size()) #define all(x) (x).begin(), (x).end() #define rall(x) (x).rbegin(), (x).rend() #define make_unique(v) sort(all(v)), v.erase(unique(all(v)), (v).end()) #define pb push_back #define mp(a, b) make_pair(a, b) #define fi first #define se second #define cmin(a, b) a = min(a, b) #define cmax(a, b) a = max(a, b) #define YES cout << "YES" << endl #define NO cout << "NO" << endl #define QYES quit("YES") #define QNO quit("NO") #define int long long #define double long double const int inf = numeric_limits<int>::max() / 4; mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); const int MOD = 1e9+7, LOG=20; void solve(); signed main() { ios::sync_with_stdio(0); cin.tie(0); int tt=1; // cin >> tt; while(tt--) solve(); return 0; } int n, k; vector<int> a, pref; vector<vector<int>> dp; int intersect(int i, int j) { return (dp[0][j] - pref[j]*pref[j] - (dp[0][i] - pref[i]*pref[i])) / (pref[i]-pref[j]); } bool better(int i, int j, int x) { return (dp[0][i] - pref[i]*pref[i]) + pref[i]*pref[x] >= (dp[0][j] - pref[j]*pref[j]) + pref[j]*pref[x]; } void solve() { read(n, k); a.clear(), a.resize(n); read(a); pref.resize(n); pref[0]=a[0]; for (int i=1; i<n; i++) pref[i]=pref[i-1]+a[i]; dp.resize(2, vector<int>(n, 0)); vector<vector<signed>> prec(k+1, vector<signed>(n, -1)); deque<int> dq; for (int i=1; i<=k; i++) { dq.clear(); dq.push_back(0); for (int j=1; j<n; j++) { while (sz(dq)>1 && better(dq[1], dq[0], j)) dq.pop_front(); dp[1][j]=dp[0][dq[0]] + pref[dq[0]] * (pref[j]-pref[dq[0]]); dbg(i, j, dp[1][j], dq); prec[i][j]=dq[0]; while (sz(dq)>1 && (pref[j]==pref[dq[sz(dq)-1]] || intersect(dq[sz(dq)-1], dq[sz(dq)-2]) >= intersect(dq[sz(dq)-1], j))) dq.pop_back(); dq.push_back(j); } swap(dp[0], dp[1]); } cout << dp[0][n-1] << endl; vector<int> res; while (k) { int p=prec[k][n-1]; res.pb(p+1); k--, n=p+1; } reverse(all(res)); print(res); } /* dp[i][j] --> i=index, j=number of segments used dp[i][j] = max(dp[k][j-1]+sum(0..k)*sum(0..i) - sum(0..k)^2) for k<i */
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