제출 #1106900

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1106900		InvMOD	Split the sequence (APIO14_sequence)	C++14	100 / 100	1577 ms	81316 KiB

sequence

#include <bits/stdc++.h>
using namespace std;

#define fi first
#define se second
#define gcd __gcd
#define sz(v) (int) v.size()
#define pb push_back
#define pi pair<int,int>
#define all(v) (v).begin(), (v).end()
#define compact(v) (v).erase(unique(all(v)), (v).end())
#define FOR(i, a, b) for(int i = (a); i <= (b); i++)
#define dbg(x) "[" #x " = " << (x) << "]"
///#define int long long

using ll = long long;
using ld = long double;
using ull = unsigned long long;

template<typename T> bool ckmx(T& a, const T& b){if(a <= b) return a = b, true; return false;}
template<typename T> bool ckmn(T& a, const T& b){if(a >= b) return a = b, true; return false;}

const int N = 1e5+1;
const int K = 201;
const ll INF = 1e18;


struct Line{
    ld a,b; int id;
    Line(ld a = 0, ld b = 0, int id = 0): a(a), b(b), id(id) {}

    ld intersect(Line& x){return (x.b - b)/(a - x.a);}
    bool operator < (const Line& x) const{return a < x.a;}
};

struct CHT{
    deque<pair<Line,ld>> dq;

    void ins(Line x){
        while(!dq.empty()){
            if(dq.back().first.a != x.a){
                if(dq.back().second >= dq.back().first.intersect(x)) dq.pop_back();
                else{
                    break;
                }
            }
            else{
                if(dq.back().first.b <= x.b) dq.pop_back();
                else{
                    return;
                }
            }
        }

        if(dq.empty()){
            dq.push_back(make_pair(x, -INF));
        }
        else dq.push_back(make_pair(x, dq.back().first.intersect(x)));
        return;
    }

    Line get(ll x){
        while(dq.size() > 1 && dq[1].second <= x) dq.pop_front();

        dq.front().second = -INF;
        return dq.front().first;
    }

    void rest(){while(dq.size()) dq.pop_back();}
};

int n, k, a[N], trace[K][N];
ll dp[2][N];

void solve()
{
    cin >> n >> k;
    for(int i = 1; i <= n; i++){
        cin >> a[i];
        a[i] += a[i-1];
    }

    CHT hull;
    for(int i = 1; i <= k; i++){
        for(int j = 1; j <= n; j++){
            hull.ins(Line(a[j-1], dp[0][j-1] - 1ll*a[j-1] * a[j-1], j));
            Line x = hull.get(a[j]);

            ll aL = x.a, bL = x.b;
            dp[1][j] = aL * (1ll*a[j]) + bL;
            trace[i][j] = x.id;
        }
        swap(dp[0], dp[1]);
        hull.rest();
    }

    cout << dp[0][n] <<"\n";

    vector<int> path;
    for(int cur_k = k, cur_pos = n; cur_k; cur_pos = trace[cur_k][cur_pos]-1, cur_k--){
        path.push_back(trace[cur_k][cur_pos]-1);
    }

    reverse(all(path));

    for(int i = 0; i < sz(path); i++) cout << path[i] <<" "; cout <<"\n";

    return;
}

signed main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    #define name "InvMOD"
    if(fopen(name".INP", "r")){
        freopen(name".INP","r",stdin);
        freopen(name".OUT","w",stdout);
    }

    int t = 1; //cin >> t;
    while(t--) solve();
    return 0;
}

컴파일 시 표준 에러 (stderr) 메시지

sequence.cpp: In function 'void solve()':
sequence.cpp:106:5: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
  106 |     for(int i = 0; i < sz(path); i++) cout << path[i] <<" "; cout <<"\n";
      |     ^~~
sequence.cpp:106:62: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
  106 |     for(int i = 0; i < sz(path); i++) cout << path[i] <<" "; cout <<"\n";
      |                                                              ^~~~
sequence.cpp: In function 'int main()':
sequence.cpp:119:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  119 |         freopen(name".INP","r",stdin);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~
sequence.cpp:120:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  120 |         freopen(name".OUT","w",stdout);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~

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