Submission #45725

#	Time	Username	Problem	Language	Result	Execution time	Memory
45725		JohnTitor	Split the sequence (APIO14_sequence)	C++11	71 / 100	2084 ms	83592 KiB

sequence

#include <bits/stdc++.h>
using namespace std;
#define FOR(i, j, k) for(int i=(j); i<=(k); i++)
#define FFOR(i, j, k) for(int i=(j); i<(k); i++)
#define DFOR(i, j, k) for(int i=(j); i>=(k); i--)
#define bug(x) cerr<<#x<<" = "<<(x)<<'\n'
#define pb push_back
#define mp make_pair
#define setbit(s, i) (s|=(1LL<<(i)))
#define bit(s, i) (((s)>>(i))&1LL)
#define mask(i) ((1LL<<(i)))
#define builtin_popcount __builtin_popcountll
typedef long long ll;
typedef long double ld;
template <typename T> inline void read(T &x){
    char c;
    bool nega=0;
    while((!isdigit(c=getchar()))&&(c!='-'));
    if(c=='-'){
        nega=1;
        c=getchar();
    }
    x=c-48;
    while(isdigit(c=getchar())) x=x*10+c-48;
    if(nega) x=-x;
}
template <typename T> inline void writep(T x){
    if(x>9) writep(x/10);
    putchar(x%10+48);
}
template <typename T> inline void write(T x){
    if(x<0){
        putchar('-');
        x=-x;
    }
    writep(x);
}
template <typename T> inline void writeln(T x){
    write(x);
    putchar('\n');
}
#define taskname "sequence"
#define sqr(x) ((x)*(x))
int n, k;
ll a[100001];
int s[100001];
ll sqs[100001];
ll f[100001];
int trace[100001][201];
deque <int> q;
const ll inf=mask(60);
class data{
public:
    int a, first;
    ll b;
//    data (int _a, ll _b, int _first){
//        a=_a;
//        b=_b;
//        first=_first;
//    }
    ll first_better(data &L){///a>=L.a
        if(a==L.a){
            if(L.b<=b) return 0;
            else return inf;
        }
        else{
            if(L.b<=b) return 0;
            return (L.b-b-1)/(a-L.a)+1;
        }
    }
};
void track(int nn, int k){
    if(nn==0) return;
    track(trace[nn][k], k-1);
    if(nn==n) return;
    write(nn);
    putchar(' ');
}
data d[100001];
int main(){
    #ifdef Kanikou
        if(fopen(taskname".inp", "r"))
            freopen(taskname".inp", "r", stdin);
    #endif // Kanikou
    read(n);
    read(k);
    FOR(i, 1, n) read(a[i]);
    FOR(i, 1, n) s[i]=s[i-1]+a[i];
    FOR(i, 1, n) sqs[i]=((ll)s[i])*s[i];
    FOR(i, 1, n) d[i].a=s[i];
    ll a;
    FOR(j, 1, k){
        q.clear();
        d[0].a=d[0].b=d[0].first=0;
        q.pb(0);
        FOR(i, 1, n){
            d[i].b=f[i]-sqs[i];
            while((q.size()>1)&&(d[q[1]].first<=s[i])) q.pop_front();
            f[i]=((ll)s[i])*d[q.front()].a+d[q.front()].b;
            trace[i][j]=q.front();
            while(true){
                a=d[i].first_better(d[q.back()]);
                if(a>=s[n]) break;
                if(a<=d[q.back()].first){
                    q.pop_back();
                    if(q.empty()){
                        d[i].first=0;
                        q.pb(i);
                        break;
                    }
                }
                else{
                    d[i].first=a;
                    q.pb(i);
                    break;
                }
            }
        }
    }
    writeln(f[n]);
    track(n, k);
}

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