Submission #391342

#	Time	Username	Problem	Language	Result	Execution time	Memory
391342		osaaateiasavtnl	Split the sequence (APIO14_sequence)	C++14	100 / 100	1368 ms	88388 KiB

sequence

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
#include <fstream>
using namespace std;
#define int long long
#define ii pair <int, int>
#define app push_back
#define all(a) a.begin(), a.end()
#define bp __builtin_popcountll
#define ll long long
#define mp make_pair
#define x first
#define y second
#define Time (double)clock()/CLOCKS_PER_SEC
#define debug(x) std::cerr << #x << ": " << x << '\n';
#define FOR(i, n) for (int i = 0; i < n; ++i)
#define FORN(i, n) for (int i = 1; i <= n; ++i)
#define pb push_back
#define trav(a, x) for (auto& a : x)
using vi = vector<int>;
template <typename T>
std::istream& operator >>(std::istream& input, std::vector<T>& data)
{
    for (T& x : data)
        input >> x;
    return input;
}
template <typename T>
std::ostream& operator <<(std::ostream& output, const pair <T, T> & data)
{
    output << "(" << data.x << "," << data.y << ")";
    return output;
}
template <typename T>
std::ostream& operator <<(std::ostream& output, const std::vector<T>& data)
{
    for (const T& x : data)
        output << x << " ";
    return output;
}
ll div_up(ll a, ll b) { return a/b+((a^b)>0&&a%b); } // divide a by b rounded up
ll div_down(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // divide a by b rounded down 
ll math_mod(ll a, ll b) { return a - b * div_down(a, b); }
#define tcT template<class T
#define tcTU tcT, class U
tcT> using V = vector<T>; 
tcT> void re(V<T>& x) { 
    trav(a, x)
        cin >> a;
}
tcT> bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0; 
} // set a = min(a,b)
tcT> bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0; 
}
ll gcd(ll a, ll b) {
    while (b) {
        tie(a, b) = mp(b, a % b);
    }
    return a;
}
 
struct Line {
    ll k, b, i;
 
    Line() : k(), b(), i() {}
    Line (ll _k, ll _b, ll _i) : k(_k), b(_b), i(_i) {}
 
    ll getVal(ll x) {
        return k * x + b;
    }
};
struct Hull {
    vector<Line> lines;
    vector<ll> borders;
 
    Hull() : lines(), borders() {}
 
    void addLine(Line L) {
        while(!lines.empty()) {
            if (lines.back().getVal(borders.back()) >= L.getVal(borders.back())) {
                lines.pop_back();
                borders.pop_back();
            } else break;
        }
        if (lines.empty()) {
            lines.push_back(L);
            borders.push_back(0LL); //leftmost query
            return;
        }
        if (lines.back().k <= L.k) return;
        ll x = div_up(L.b - lines.back().b, lines.back().k - L.k); //must work for negative!
        lines.push_back(L);
        borders.push_back(x);
    }
    int getMinVal(ll x) {
        int pos = upper_bound(borders.begin(), borders.end(), x) - borders.begin();
        if (pos == 0) return -1;
        pos--;
        return lines[pos].i;
    }
};
 
const int N = 1e5+7, K = 207, INF = 1e18;
signed par[K][N];
int dp1[N], dp2[N];
 
signed main() {
    #ifdef LOCAL
    #else
    #define endl '\n'
    ios_base::sync_with_stdio(0); cin.tie(0);
    #endif
    int n, k;
    cin >> n >> k;
    k++;
    vi a(n);
    cin >> a;
    vi p(n + 1);
    FOR (i, n) {
    	p[i + 1] = p[i] + a[i];
    }
 
    FOR (i, N) {
    	dp1[i] = INF;
    }
    dp1[0] = 0;
 
    FORN (t, k) {
    	Hull h;
    	FOR (i, n + 1) {
    		int j = h.getMinVal(p[i] * 2);
    		if (j != -1) {
    			dp2[i] = dp1[j] + (p[i] - p[j]) * (p[i] - p[j]); 
    			par[t][i] = j;   		
    		}
    		else {
    			dp2[i] = INF;
    		}
    		if (dp1[i] != INF) {
    			h.addLine(Line(-p[i], dp1[i] + p[i] * p[i], i));
    		}
    	}
    	FOR (i, N) {
    		dp1[i] = dp2[i];
    	}
    }
    
 
    cout << (p[n] * p[n] - dp1[n]) / 2 << endl;
    vi cut;
    int i = n;
    for (int t = k; t; --t) {
    	cut.app(i);
    	i = par[t][i];
    }
    reverse(all(cut));
    cut.pop_back();
    cout << cut << endl;
}

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