Submission #311366

#	Time	Username	Problem	Language	Result	Execution time	Memory
311366		zaneyu	Split the sequence (APIO14_sequence)	C++14	100 / 100	1322 ms	96344 KiB

sequence

/*input
7 3
4 1 3 4 0 2 3
*/
#include<bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
//using namespace __gnu_pbds;
//typedef tree<pair<int,int>, null_type, less<pair<int,int>>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
typedef long long int ll;
#define ld double
#define pii pair<int,int>
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(ll i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
const ll maxn=1e5+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=3.14159265358979323846;
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#define MP make_pair
ll mult(ll a,ll b){
    return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b){
    ll res=1LL;
    while(b){
        if(b&1) res=mult(res,a);
        a=mult(a,a);
        b>>=1;
    }
    return res;
}
struct line{
    ll m,c;
    ll get(ll x){
        return x*m+c;
    }
    ld intersect(line x){
        return (ld)(x.c-c)/(m-x.m+0.000000001);
    }
};
int par[maxn][205];
deque<pair<line,int>> dq[205];
ll pref[maxn];
 
int main(){
    ios::sync_with_stdio(false),cin.tie(0);
    int n,k;
    cin>>n>>k;
    REP1(i,n){
        int x;
        cin>>x;
        pref[i]=pref[i-1]+x;
    }
    REP(j,k){
        line c={0,0};
        dq[j].pb({c,0});
    }
    ll ans=0;
    REP1(i,n){
        for(int j=k;j>=1;j--){
            while(sz(dq[j-1])>1 and dq[j-1].front().f.get(pref[i])<=dq[j-1][1].f.get(pref[i])){
                dq[j-1].pop_front();
            }
            //REP(z,i) MXTO(dp[i][j],dp[z][j-1]+(pref[i]-pref[z])*pref[z]);
            ll dp=dq[j-1].front().f.get(pref[i]);
            //cout<<dp<<'\n';
            par[i][j]=dq[j-1].front().s;
            MXTO(ans,dp);
            //cout<<i<<' '<<j<<' '<<dp[i][j]<<'\n';
            line c={pref[i],dp-pref[i]*pref[i]};
            while(sz(dq[j])>1 and c.intersect(dq[j].back().f)<=dq[j].back().f.intersect(dq[j][sz(dq[j])-2].f)) dq[j].pop_back();
            dq[j].pb({c,i});
        }
        line c={pref[i],-pref[i]*pref[i]};
        dq[0].pb({c,i});
    }
    cout<<ans<<'\n';
    int ci=n,cj=k;
    vector<int> v;
    while(cj){
        ci=par[ci][cj];
        v.pb(ci);
        --cj;
    }
    reverse(ALL(v));
    for(auto x:v) cout<<x<<' ';
}

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