Submission #971241

#	Time	Username	Problem	Language	Result	Execution time	Memory
971241		Unforgettablepl	Split the sequence (APIO14_sequence)	C++17	100 / 100	593 ms	83932 KiB

sequence

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

#define int long long

struct fraction {
    int n,d;
    fraction(int n,int d):n(n),d(d){}
    bool operator<(const fraction &other) const {
        return __int128(n)*__int128(other.d)<__int128(other.n)*__int128(d);
    }
};

struct line {
    int m,k,idx_1,idx_2;
    line(int m,int k,int idx_1,int idx_2):m(m),k(k),idx_1(idx_1),idx_2(idx_2){}
    int operator()(int x) const {
        return m*x+k;
    }
};

fraction getIntersect(line a,line b) {
    return {a.k - b.k,b.m - a.m};
}

deque<line> hull;

void addline(line x) {
    if(!hull.empty() and hull.back().m==x.m){
        if(hull.back().k<=x.k){
            hull.pop_back();
            addline(x);
        }
        return;
    }
    while(hull.size()>1 and getIntersect(hull[hull.size()-2],x)<getIntersect(hull[hull.size()-2],hull.back()))hull.pop_back();
    hull.emplace_back(x);
}
tuple<int,int,int> get(int x){
    while(hull.size()>=2 and hull.front()(x)<=hull[1](x))hull.pop_front();
    return {hull.front()(x),hull.front().idx_1,hull.front().idx_2};
}

int DP[2][100001];
int32_t back[100001][202];
int P[100001];

int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n,k;
    cin >> n >> k;k++;
    for(int i=1;i<=n;i++)cin>>P[i];
    for(int i=1;i<=n;i++)P[i]+=P[i-1];
    for(int i=1;i<=n;i++)DP[1][i]=-1e18;
    for(int j=1;j<=k;j++){
        swap(DP[1],DP[0]);
        hull.clear();
        addline({0,0,0,0});
        for(int i=1;i<=n;i++){
            auto [ans,last1,last2] = get(P[i]);
            DP[1][i] = ans;
            back[i][j] = last1;
            addline({P[i],DP[0][i]-P[i]*P[i],i,j-1});
        }
    }
    cout << DP[1][n] << '\n';
    vector<int> ans;
    auto curr1 = back[n][k];
    int32_t curr2 = k-1;
    while(curr1 and curr2){
        ans.emplace_back(curr1);
        curr1 = back[curr1][curr2];
        curr2--;
    }
    for(int&i:ans)cout<<i<<' ';
    cout << '\n';
}

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