| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 367510 | arnold518 | Split the sequence (APIO14_sequence) | C++14 | 325 ms | 7524 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int MAXN = 1e5;
int N, K;
ll A[MAXN+10];
struct Line
{
ll a, b; int p;
};
double cross(Line p, Line q) { return (double)(p.b-q.b)/(q.a-p.a); }
struct CHT
{
vector<Line> S;
void init()
{
S.clear();
}
void update(Line p)
{
if(!S.empty() && S.back().a==p.a)
{
if(S.back().b>p.b)
{
S.pop_back();
S.push_back(p);
}
return;
}
while(S.size()>1 && cross(S[S.size()-2], S[S.size()-1])>=cross(S[S.size()-1], p)) S.pop_back();
S.push_back(p);
}
int query(ll x)
{
int lo=0, hi=S.size();
while(lo+1<hi)
{
int mid=lo+hi>>1;
if(cross(S[mid], S[mid-1])<=x) lo=mid;
else hi=mid;
}
return S[lo].p;
}
}cht;
pll dp[MAXN+10];
int memo[MAXN+10];
vector<int> V, ansV;
pll solve(ll lambda)
{
cht.init();
cht.update({0, 0, 0});
V.clear();
for(int i=1; i<=N; i++)
{
int t=cht.query(A[i]);
dp[i].first=dp[t].first+(A[i]-A[t])*(A[i]-A[t])*2+lambda;
dp[i].second=dp[t].second+1;
memo[i]=t;
cht.update({-2*A[i], A[i]*A[i]+dp[i].first, i});
}
for(int i=N; i>0;)
{
V.push_back(i);
i=memo[i];
}
V.push_back(0);
reverse(V.begin(), V.end());
for(auto it : V)
//printf("%lld : %lld %lld\n", lambda, dp[N].first, dp[N].second);
return dp[N];
}
int main()
{
scanf("%d%d", &N, &K); K++;
for(int i=1; i<=N; i++) scanf("%lld", &A[i]);
for(int i=1; i<=N; i++) A[i]+=A[i-1];
ll lo=0, hi=1e9;
while(lo+1<hi)
{
ll mid=lo+hi>>1;
if(solve(2*mid+1).second<=K) hi=mid;
else lo=mid;
}
vector<int> L, R;
solve(2*hi+1); L=V;
solve(2*lo+1); R=V;
if(L.size()-1==K)
{
ansV=L;
}
else if(R.size()-1==K)
{
ansV=R;
}
else
{
assert(L.size()<K && K<R.size());
for(int i=0; i+1<R.size(); i++)
{
int l=R[i], r=R[i+1];
int lt=upper_bound(L.begin(), L.end(), l)-L.begin()-1;
if(L[lt]<=l && r<=L[lt+1] && i+L.size()-lt-1==K)
{
for(int j=0; j<=i; j++) ansV.push_back(R[j]);
for(int j=lt+1; j<R.size(); j++) ansV.push_back(L[j]);
break;
}
}
}
ll ans=solve(2*hi).first/2-hi*K;
ans=(A[N]*A[N]-ans)/2;
printf("%lld\n", ans);
for(int i=1; i+1<ansV.size(); i++) printf("%d ", ansV[i]);
}Compilation message (stderr)
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