Submission #1101514

#TimeUsernameProblemLanguageResultExecution timeMemory
1101514doducanhSplit the sequence (APIO14_sequence)C++14
0 / 100
11 ms5456 KiB
///breaker #pragma GCC optimize("O2") #include<bits/stdc++.h> using namespace std; #define int long long #define ll long long #define fi first #define se second #define ii pair<int,int> #define mp make_pair #define in(x) freopen(x,"r",stdin) #define out(x) freopen(x,"w",stdout) #define bit(x,i) ((x>>i)&1) #define lc (id<<1) #define rc ((id<<1)^1) const int maxn=1e5+7; ll dp[2][maxn]; int where[maxn][205]; ll a[maxn]; int n,k; double slope(int i,int j) { return (double)(dp[0][i]-dp[0][j])/(a[i]-a[j]); } void sol(void){ cin>>n>>k; for(int i=1;i<=n;i++){ cin>>a[i]; a[i]+=a[i-1]; } for(int i=0;i<=n;i++){ dp[0][i]=dp[1][i]=0; } deque<int>dq; for(int i=1;i<=k;i++){ dq.push_back(i-1); for(int j=i;j<=n;j++){ while(dq.size()>1&&a[n]-a[j]<=slope(dq[0],dq[1]))dq.pop_front(); int tmp=dq.front(); where[i][j]=tmp; dp[1][j]=dp[0][tmp]+1ll*a[j]*(a[n]-a[j])-1ll*a[tmp]*(a[n]-a[j]); while(dq.size()>1&&slope(dq[dq.size()-2],dq.back())>=slope(dq.back(),j))dq.pop_back(); dq.push_back(j); } dq.clear(); for(int j=i;j<=n;j++)dp[0][j]=dp[1][j]; } int indx=k; for(int i=k+1;i<=n;i++){ if(dp[0][i]>dp[0][indx])indx=i; } vector<int>v; cout<<dp[0][indx]<<"\n"; for(int i=k;i>=1;i--){ v.push_back(indx); indx=where[i][indx]; } reverse(v.begin(),v.end()); for(int x:v)cout<<x<<" "; } signed main(){ ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0); int t=1; while(t--){ sol(); } // you should actually read the stuff at the bottom return 0; } /* stuff you should look for * int overflow, array bounds * special cases (n=1?) * do smth instead of nothing and stay organized * WRITE STUFF DOWN * DON'T GET STUCK ON ONE APPROACH */
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