Submission #569127

#	Time	Username	Problem	Language	Result	Execution time	Memory
569127		n0sk1ll	Split the sequence (APIO14_sequence)	C++14	49 / 100	141 ms	131072 KiB

sequence

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

#define FAST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);cerr.tie(0)
#define mp make_pair
#define xx first
#define yy second
#define pb push_back
#define pf push_front
#define popb pop_back
#define popf pop_front
#define all(x) (x).begin(),(x).end()
#define erase_uni(x) x.erase(unique(all(x)),x.end())
#define inv(n) power((n), mod - 2)
#define ff(i,a,b) for (int (i) = (a); (i) < (b); (i)++)
#define fff(i,a,b) for (int (i) = (a); (i) <= b; (i)++)
#define bff(i,a,b) for (int (i) = (b)-1; (i) >= (a); (i)--)
#define bfff(i,a,b) for (int (i) = (b); (i) >= (a); (i)--)
#define sum_overflow(a,b) __builtin_add_overflow_p ((a), (b), (__typeof__ ((a) + (b))) 0)
#define mul_overflow(a,b) __builtin_mul_overflow_p ((a), (b), (__typeof__ ((a) + (b))) 0)

using namespace std;
long double typedef ld;
unsigned int typedef ui;
long long int typedef li;
pair<int,int> typedef pii;
pair<li,li> typedef pli;
pair<ld,ld> typedef pld;
vector<vector<int>> typedef graph;
unsigned long long int typedef ull;
//const int mod = 998244353;
const int mod = 1000000007;

/*using namespace __gnu_pbds;
template <class T> using oset = tree<T, null_type,less<T>, rb_tree_tag,tree_order_statistics_node_update>;
template <class T> using omset = tree<T, null_type,less_equal<T>, rb_tree_tag,tree_order_statistics_node_update>;*/







//Note to self: Check for overflow

int a[100005];
li dp[100005][202];
int gde[100005][202];
int pre[100005];

li cost(int l, int r)
{
    if (l==1) return pre[r];
    return (li)pre[l-1]*(pre[r]-pre[l-1]);
}

void compute(int k, int l, int r, int optl, int optr)
{
    if (l>r) return;
    int mid=(l+r)/2;
    fff(g,optl,min(mid-1,optr))
    {
        li sta=dp[g][k-1]+cost(g+1,mid);
        if (sta>=dp[mid][k]) dp[mid][k]=sta,gde[mid][k]=g;
    }

    compute(k,l,mid-1,optl,gde[mid][k]);
    compute(k,mid+1,r,gde[mid][k],optr);
}

void rekurz(int n, int k)
{
    if (k==0) return;
    rekurz(gde[n][k],k-1),cout<<max(1,gde[n][k])<<" ";
}

int main()
{
    FAST;

    int n,k; cin>>n>>k;
    fff(i,1,n) cin>>a[i];

    fff(i,1,n) pre[i]=pre[i-1]+a[i];
    fff(j,1,k) compute(j,1,n,1,n);



    /*fff(j,0,k)
    {
        cout<<j<<": "<<endl;
        fff(i,1,n) cout<<dp[i][j]<<" "; cout<<endl;
        fff(i,1,n) cout<<gde[i][j]<<" "; cout<<endl<<endl;
    }*/

    cout<<dp[n][k]<<"\n";
    if (n<=10000) rekurz(n,k);
}

//Note to self: Check for overflow

/*

7 3
4 1 3 4 0 2 3

7 6
4 1 3 4 0 2 3

*/

Compilation message (stderr)

sequence.cpp: In function 'void compute(int, int, int, int, int)':
sequence.cpp:17:29: warning: unnecessary parentheses in declaration of 'g' [-Wparentheses]
   17 | #define fff(i,a,b) for (int (i) = (a); (i) <= b; (i)++)
      |                             ^
sequence.cpp:62:5: note: in expansion of macro 'fff'
   62 |     fff(g,optl,min(mid-1,optr))
      |     ^~~
sequence.cpp: In function 'int main()':
sequence.cpp:17:29: warning: unnecessary parentheses in declaration of 'i' [-Wparentheses]
   17 | #define fff(i,a,b) for (int (i) = (a); (i) <= b; (i)++)
      |                             ^
sequence.cpp:83:5: note: in expansion of macro 'fff'
   83 |     fff(i,1,n) cin>>a[i];
      |     ^~~
sequence.cpp:17:29: warning: unnecessary parentheses in declaration of 'i' [-Wparentheses]
   17 | #define fff(i,a,b) for (int (i) = (a); (i) <= b; (i)++)
      |                             ^
sequence.cpp:85:5: note: in expansion of macro 'fff'
   85 |     fff(i,1,n) pre[i]=pre[i-1]+a[i];
      |     ^~~
sequence.cpp:17:29: warning: unnecessary parentheses in declaration of 'j' [-Wparentheses]
   17 | #define fff(i,a,b) for (int (i) = (a); (i) <= b; (i)++)
      |                             ^
sequence.cpp:86:5: note: in expansion of macro 'fff'
   86 |     fff(j,1,k) compute(j,1,n,1,n);
      |     ^~~

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