제출 #406789

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
406789		JerryLiu06	수열 (APIO14_sequence)	C++17	71 / 100	82 ms	131076 KiB

sequence

#include <bits/stdc++.h>
 
using namespace std;
 
using ll = long long;
using db = long double;
using str = string;
 
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
 
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
 
#define mp make_pair
#define f first
#define s second
 
#define sz(x) (int)(x).size()
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define sor(x) sort(all(x))
#define ft front()
#define bk back()
#define pb push_back
#define pf push_front
 
#define lb lower_bound
#define ub upper_bound
 
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) FOR(i, 0, a)
#define ROF(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define R0F(i, a) ROF(i, 0, a)
#define EACH(a, x) for (auto& a : x)
 
const int MOD = 1e9 + 7;
const int MXN = 1e5 + 10;
const int MXK = 210;
const ll INF = 1e18;
 
using ld = long double;
 
int N, K; ll S[MXN], DP[MXN][2], ans[MXN][MXK];
 
ld isect(int A, int B, int i) {
    return (ld) (S[A] * S[A] + DP[A][i] - S[B] * S[B] - DP[B][i]) / (ld) (2 * S[A] - 2 * S[B]); 
}

bool check1(int A, int B, int i, int j) {
    return S[A] * S[A] + DP[A][j] - S[B] * S[B] - DP[B][j] >= S[i] * (2 * S[A] - 2 * S[B]);
}
bool check2(int A, int B, int i, int j) {
    return (S[A] * S[A] + DP[A][j] - S[B] * S[B] - DP[B][j]) * (2 * S[B] - 2 * S[i]) >=
           (S[B] * S[B] + DP[B][j] - S[i] * S[i] - DP[i][j]) * (2 * S[A] - 2 * S[B]);
}
 
int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);
 
    cin >> N >> K; FOR(i, 1, N + 1) { int A; cin >> A; S[i] = A + S[i - 1]; }
 
    FOR(i, 1, N + 1) DP[i][1] = S[i] * S[i];
 
    FOR(j, 2, K + 2) {
        deque<int> Q; Q.pb(0); FOR(i, 1, N + 1) {
          //  if (sz(Q) > 1) cout << isect(Q[0], Q[1], (j - 1) % 2) << endl; 
            
            while (sz(Q) > 1 && check1(Q[0], Q[1], i, (j - 1) % 2)) Q.pop_front();
 
            int cur = Q.front(); DP[i][j % 2] = DP[cur][(j - 1) % 2] + (S[i] - S[cur]) * (S[i] - S[cur]); ans[i][j] = cur;
            
     //       cout << cur << " " << DP[i][j % 2] << "\n";
 
            while (sz(Q) > 1 && check2(Q[sz(Q) - 2], Q.back(), i, (j - 1) % 2)) Q.pop_back(); Q.pb(i);
        }
    }
    //FOR(i, 1, N + 1) cout << DP[i][1] << endl;
    
    cout << (S[N] * S[N] - DP[N][(K + 1) % 2]) / 2 << "\n"; int ind = N;
 
    ROF(i, 2, K + 2) { ind = ans[ind][i]; cout << (!ind ? 1 : ind) << " "; }
}

컴파일 시 표준 에러 (stderr) 메시지

sequence.cpp: In function 'int main()':
sequence.cpp:82:13: warning: this 'while' clause does not guard... [-Wmisleading-indentation]
   82 |             while (sz(Q) > 1 && check2(Q[sz(Q) - 2], Q.back(), i, (j - 1) % 2)) Q.pop_back(); Q.pb(i);
      |             ^~~~~
sequence.cpp:82:95: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'while'
   82 |             while (sz(Q) > 1 && check2(Q[sz(Q) - 2], Q.back(), i, (j - 1) % 2)) Q.pop_back(); Q.pb(i);
      |                                                                                               ^

• Subtask 1: 11 / 11
• Subtask 2: 11 / 11
• Subtask 3: 11 / 11
• Subtask 4: 17 / 17
• Subtask 5: 21 / 21
• Subtask 6: 0 / 29