Submission #1202546

#	Time	Username	Problem	Language	Result	Execution time	Memory
1202546		Faisal_Saqib	Split the sequence (APIO14_sequence)	C++20	0 / 100	0 ms	328 KiB

sequence

/*
    VENI VIDI VICI
*/

// #pragma GCC optimize("Ofast","unroll-all-loops","fast-math","no-stack-protector","give-no-fuck")
#include <bits/stdc++.h>
// #include <iostream>
// #include <vector>
// #include <algorithm>
// #include <set>
// #include <map>

using namespace std;
#define fi first
#define se second
#define pb push_back
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()

#define rep(i,x, n) for (int i = (x); i < (n); ++i)
#define repr(i,x, n) for (int i = (n) - 1; i >= (x); --i)
// #define sum accumulate

//using i128 = __int128;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;


using vi = vector<int>;
using vl = vector<ll>;
using vpi = vector<pair<int, int>>;
using vvi = vector<vi>;
using sll = set<ll>;

template<class T>
istream& operator>>(istream& is, vector<T>& v) {
    for(auto &i:v) is>>i;
    return is;
}
template<class T1,class T2>
istream& operator>>(istream& is, pair<T1,T2>& p) {
    is>>p.fi>>p.se;
    return is;
}
template<class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
    for(const auto &i:v) os<<i<<' ';
    return os;
}
template<class T1,class T2>
ostream& operator<<(ostream& os, const pair<T1,T2>& p) {
    os<<p.fi<<' '<<p.se; return os;
}
void pyn(bool x)
{
    cout<<(x?"YES":"NO")<<endl;
}
void pYN(bool x)
{
    cout<<(x?"Yes":"No")<<endl;
}
void pAB(bool x)
{
    cout<<(x?"Alice":"Bob")<<endl;
}
ll powmod(ll a,ll b,ll modulo)
{
  if(b==0){
    return 1;
  }
  ll temp=powmod(a,b/2,modulo);
  if(b%2==0){
    return (temp*temp)%modulo;
  }
  else{
    return (a*((temp*temp)%modulo))%modulo;
  }
}
bool Prime(ll n){
    for (ll i = 2; i*i <= n; i++)
        if (n % i == 0)
            return false;
    return (n>1);
}
const ll N=1e5+1,K=201,inf=1e18;
struct line
{
    ld m,c;
    int ind;
    line(ld x,ld y,int opt)
    {
        m=x;
        c=y;
        ind=opt;
    }
    pair<ld,int> get(ld x)
    {
        return {m*x+c,ind};
    }
    ld inter(const line&b)
    {
      if(m==b.m)
      {
        if(c>b.c)
        {
            return inf;
        }
        else
        {
          return -inf;
        }
      }
        return (b.c-c)/(m-b.m);
    }
};

ll n,k;
ll a[N],pre[N];
ll dp[K][N];
int bk[K][N];
deque<line> dsa;
void insert(ld x,ld y,int opt)
{
    line cur(x,y,opt);
    int sz=dsa.size();
    while(dsa.size()>=2 and dsa[sz-2].inter(dsa[sz-1])>=dsa[sz-2].inter(cur))
    {
        dsa.pop_back();
        sz--;
    }
    dsa.pb(cur);
}
pair<ld,int> getans(ld v) // queries are increasing so just pop front
{
    int s=0,e=dsa.size();
    // while(dsa.size()>=2 and dsa[0].inter(dsa[1])<v)
    // {
    //     dsa.pop_front();
    // }
    // return dsa[0].get(v);
    while(s+1<e)
    {
        int mid=(s+e)/2;
        if(dsa[mid-1].inter(dsa[mid])>v)
        {
            e=mid;
        }
        else
        {
            s=mid;
        }
    }
    return dsa[s].get(v);
}
void solve()
{
    cin>>n>>k;
    k++;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        pre[i]=pre[i-1]+a[i];
    }
    for(int i=0;i<=k;i++)
    {
        for(int j=0;j<=n;j++)
        {
            dp[i][j]=-inf;
            if(i==0)dp[i][j]=0;
        }
    }
    // dp[0][0]=0;
    for(int i=1;i<=k;i++)
    {
        // c = pre[j]-pre[j']
        // s = pre[n]
        // (s-c) * c
        // (pre[n]-pre[j]+pre[j'])*(pre[j]-pre[j'])
        // pre[n]*pre[j] - pre[n]*pre[j']    + pre[j]*pre[j'] - pre[j]^2  +pre[j']*pre[j]  - pre[j']*pre[j']

        // pre[n]*pre[j] - pre[j]*pre[j] 


        // 2*pre[j']*pre[j]  - pre[j']*(pre[j']+pre[n])

        // dp[i][j] = max(dp[i-1][j']  + ); 
        insert(2*pre[0],-pre[0]*(pre[n]+pre[0]) + dp[i-1][0],0);
        for(int j=1;j<=n;j++)
        {
            auto tp=getans(pre[j]);
            dp[i][j]=tp.first + pre[n]*pre[j] - pre[j]*pre[j];
            bk[i][j]=tp.second;
            insert(2*pre[j],-pre[j]*(pre[n]+pre[j])+dp[i-1][j],j);
        }
    }
    cout<<dp[k][n]/2<<endl;
    int lp=n;
    vector<int> ans;
    for(int p=k;bk[p][lp]>0;p--)
    {
        ans.pb(bk[p][lp]);
        lp=bk[p][lp];
    }
    reverse(all(ans));
    cout<<ans<<endl;
}

int main() {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int uwu=1;

    // cin>>uwu;

    while(uwu--) {
        solve();
    }
    return 0;
}

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