제출 #1101514

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1101514		doducanh	수열 (APIO14_sequence)	C++14	0 / 100	11 ms	5456 KiB

sequence

///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
 */

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