Submission #670201

#	Time	Username	Problem	Language	Result	Execution time	Memory
670201		MohammadAghil	Split the sequence (APIO14_sequence)	C++17	100 / 100	1229 ms	84132 KiB

sequence

      #include <bits/stdc++.h>
//   #pragma GCC optimize ("Ofast,unroll-loops")
// #pragma GCC target ("avx2")
 using namespace std;
  typedef long long ll;
   typedef pair<ll, int> pp;
     #define per(i,r,l) for(int i = (r); i >= (l); i--)
       #define rep(i,l,r) for(int i = (l); i < (r); i++)
          #define all(x) begin(x), end(x)
             #define sz(x) (int)(x).size()
                 #define pb push_back
                     #define ss second
                          #define ff first
                                  void err(istringstream *iss){}template<typename T,typename ...Args> void err(istringstream *iss,const T &_val, const Args&...args){string _name;*iss>>_name;if(_name.back()==',')_name.pop_back();cerr<<_name<<" = "<<_val<<", ",err(iss,args...);}
void IOS(){
     cin.tie(0) -> sync_with_stdio(0);
     #ifndef ONLINE_JUDGE
          #define er(args ...) cerr << __LINE__ << ": ", err(new istringstream(string(#args)), args), cerr << endl
     #else
          #define er(args ...) 0
     #endif
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll mod = 1e9 + 7, maxn = 1e6 + 5, maxk = 2e2 + 5, p = 9973, lg = 22, inf = ll(1e18) + 5;
ll pw(ll a,ll b,ll md=mod){if(!b)return 1;ll k=pw(a,b>>1ll,md);return k*k%md*(b&1ll?a:1)%md;}

ll dp[maxn], tmp[maxn];
ll sum[maxn], sum2[maxn];
int opt[maxk][maxn];

ll get(int l, int r){
     ll s = sum[r] - sum[l-1];
     return s*s - sum2[r] + sum2[l-1];
} 

void slv(int l, int r, int lx, int rx, int t){
     if(l > r) return;
     int mid = (l + r)>>1;
     pp bst = {inf, -1};
     per(i,min(mid, rx), lx){
          bst = min(bst, {get(i, mid) + dp[i-1], -i});
     }
     bst.ss = -bst.ss;
     opt[t][mid] = bst.ss;
     tmp[mid] = bst.ff;
     slv(l, mid-1, lx, bst.ss, t);
     slv(mid+1, r, bst.ss, rx, t);
}

int main(){ IOS();     
     int n, k; cin >> n >> k;
     vector<ll> a(n+1);
     rep(i,1,n+1){
          cin >> a[i];
          sum[i] = sum[i-1] + a[i];
          sum2[i] = sum2[i-1] + a[i]*a[i];
     }
     rep(i,1,n+1) dp[i] = get(1, i);
     rep(i,0,k){
          slv(1, n, 1, n, i+1);
          rep(i,0,n+1) dp[i] = tmp[i];
     }
     cout << (get(1, n) - dp[n])/2ll << '\n';
     int cr = n;
     vector<int> ans;
     per(i,k,1){
          ans.pb(opt[i][cr]-1);
          cr = ans.back();
     }
     reverse(all(ans));
     for(int c: ans) cout << c << ' ';
     cout << '\n';
     return 0-0;
}

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