Submission #13761

#	Time	Username	Problem	Language	Result	Execution time	Memory
13761		comet	Split the sequence (APIO14_sequence)	C++98	0 / 100	0 ms	2776 KiB

sequence

#include<cstdio>
#include<algorithm>
#include<deque>
#include<stdlib.h>
#include<time.h>
#include<cstring>
#define MOD (ll)(1e9)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pp;
ll d[2][1010],s[100010],N,K;
short r[201][1001];
void f(int x,int y){
	if(x==1)return;
	f(x-1,r[x][y]);
	printf("%d ",(int)r[x][y]);
}
/*void f2(int x,int y){
	if(x==1)return;
	f(x-1,r2[x][y]);
	printf("%d ",(int)r2[x][y]);
}*/
struct convex{
	deque<ll> a,b,c;
	void init(){a.clear();b.clear();c.clear();}
	bool check(ll z0,ll z1,ll z2,ll z3){
		ll uz0=z0/MOD,uz1=z1/MOD,uz2=z2/MOD,uz3=z3/MOD;
		z0%=MOD,z1%=MOD,z2%=MOD,z3%=MOD;
		ll p=0,q=0,r=0,p2=0,q2=0,r2=0;
		r=z0*z1,r2=z2*z3;
		q+=r/MOD,q2+=r2/MOD;
		r%=MOD,r2%=MOD;
		q+=z0*uz1+uz0*z1,q2+=z2*uz3+z3*uz2;
		p+=q/MOD,p2+=q2/MOD;
		q%=MOD,q2%=MOD;
		p+=uz0*uz1,p2+=uz2*uz3;

		if(p!=p2)return p>p2;
		if(q!=q2)return q>q2;
		return r>r2;
	}
	void push(ll n,ll m,ll v){
		while(a.size()>1&&check(b.back()-b[b.size()-2],a.back()-n,m-b.back(),a[a.size()-2]-a.back())){
			a.pop_back();
			b.pop_back();
			c.pop_back();
		}
		a.push_back(n);
		b.push_back(m);
		c.push_back(v);
	}
	pp query(ll x){
		while(!a.empty()&&(a[0]-a[1])*x<(b[1]-b[0])){
			a.pop_front();
			b.pop_front();
			c.pop_front();
		}
		return pp(a[0]*x+b[0],c[0]);
	}
};
int main(){
	srand(time(NULL));
	scanf("%lld%lld",&N,&K);
//for(int comet=0;comet<100000;comet++){


	if(K==0)K++;
	for(int i=1;i<=N;i++){
		scanf("%lld",&s[i]);
		//s[i]=rand();
		s[i]+=s[i-1];
	}
	convex* pre=new convex;
	convex* cur=new convex;
	pp ret;
	for(int i=1;i<=N;i++){
		d[1][i]=s[i]*(s[N]-s[i]);
		pre->push(s[i],d[1][i]-s[i]*s[N],i);
	}
	for(int g=2;g<=K+1;g++){
		for(int i=g;i<=N;i++){
			ret=pre->query(s[i]);
			d[g&1][i]=ret.first+(s[N]-s[i])*s[i];
			r[g][i]=ret.second;
			cur->push(s[i],d[g&1][i]-s[i]*s[N],i);
		}
		pre->init();
		swap(cur,pre);
	}

	/*for(int i=1;i<=N;i++)d2[1][i]=s[i]*(s[N]-s[i]);
    for(int g=2;g<=K+1;g++){
        for(int i=g;i<=N;i++){
            for(int j=1;j<i;j++){
                if(d2[g%2][i]<d2[(g+1)%2][j]+(s[i]-s[j])*(s[N]-s[i])){
                    d2[g%2][i]=d2[(g+1)%2][j]+(s[i]-s[j])*(s[N]-s[i]);
                    r2[g][i]=j;
                }
            }
        }
    }*/
	/*if(d[(K+1)&1][N]!=d2[(K+1)%2][N]){
		printf("%lld %lld\n",N,K);
		for(int i=1;i<=N;i++)printf("%lld ",s[i]);
		printf("\nans= %lld, %lld\n",d[(K+1)&1][N],d2[(K+1)%2][N]);
		f(K+1,N);
		f2(K+1,N);
		break;
	}
*/
    printf("%lld \n",d[(K+1)&1][N]);
    f(K+1,N);
	//f2(K+1,N);
delete cur,pre;
}

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