Submission #900007

#	Time	Username	Problem	Language	Result	Execution time	Memory
900007		ByeWorld	Split the sequence (APIO14_sequence)	C++14	21 / 100	318 ms	131072 KiB

sequence

#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#define bupol __builtin_popcount
//#define int long long 
#define ll long long
#define ld long double
#define fi first
#define se second
#define pb push_back
#define lf (id<<1)
#define rg ((id<<1)|1)
#define md ((l+r)>>1)
using namespace std;
const int MAXN = 1e5+5;
const int MAXK = 205;
const int LOG = 20;
const int MOD = 1e9+7;
const int SQRT = 520;
const ll INF = 2e18;
typedef pair<int,int> pii;
typedef pair<int,pii> ipii;

int n, k;
int a[MAXN], pr[MAXN];
ll dp[MAXN][MAXK];

signed main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin >> n >> k;
    for(int i=1; i<=n; i++){
        cin >> a[i]; pr[i] = pr[i-1]+a[i];
    }
    for(int i=0; i<=n+1; i++)
        for(int j=1; j<=k+1; j++) dp[i][j] = -1;
            
    for(int m=1; m<=k; m++){
        int las = 1;
        for(int i=2; i<=n; i++){
            dp[i][m] = -INF; 
            int opt = -1;
            for(int j=las; j<=i-1; j++){
                if(dp[j][m-1] == -1) continue; // gk ada ke belakangnya
                if(dp[j][m-1] + 1ll * pr[j] * (pr[i]-pr[j]) > dp[i][m]){ // cek yg optimal potong dmn
                    dp[i][m] = dp[j][m-1] + 1ll * pr[j] * (pr[i]-pr[j]); // cmn ngebuild sampe dp[i][i-1]
                    opt = j;
                }
            }
            las = opt;
            //cout << dp[i][m] <<' ' << i << ' ' << m << '\n';
        }
    }
    cout << dp[n][k] << '\n';
    int nw = n;
    for(int m=k; m>=1; m--){
        int i = nw;
        dp[i][m] = -INF; 
        int opt = -1;
        for(int j=1; j<=i-1; j++){
            if(dp[j][m-1] == -1) continue; // gk ada ke belakangnya
            if(dp[j][m-1] + 1ll * pr[j] * (pr[i]-pr[j]) > dp[i][m]){ // cek yg optimal potong dmn
                dp[i][m] = dp[j][m-1] + 1ll * pr[j] * (pr[i]-pr[j]); // cmn ngebuild sampe dp[i][i-1]
                opt = j;
            }
        }
        nw = opt;
        //cout << dp[i][m] <<' ' << i << ' ' << m << '\n';
        cout << nw << ' ';
    } 
    cout << '\n';
}

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