Submission #870885

#	Time	Username	Problem	Language	Result	Execution time	Memory
870885		ngjabach	Holding (COCI20_holding)	C++14	110 / 110	76 ms	229716 KiB

holding

// NgJaBach: Forever Meadow <3

#include<bits/stdc++.h>

using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef long long int ll;
typedef unsigned long long ull;
#define pb push_back
#define pob pop_back
#define mp make_pair
#define upb upper_bound
#define lwb lower_bound
#define bend(a) a.begin(),a.end()
#define rev(x) reverse(bend(x))
#define mset(a) memset(a, 0, sizeof(a))
#define fi first
#define se second
#define gcd __gcd
#define getl(s) getline(cin, s);
#define setpre(x) fixed << setprecision(x)
#define endl '\n'
const int N=105,M=1000000007;
const ll INF=1e18+7;
int n,L,R,k,a[N];
namespace sub2{
    int dp[N][N][N*N/4];
    int solve(){
        k=min(k,n*n/4);
        for(int i=0;i<=n;++i){
            for(int j=0;j<=n;++j){
                for(int d=0;d<=k;++d){
                    dp[i][j][d]=0;
                }
            }
        }
        for(int i=1;i<L;++i){
            for(int j=L;j<=n;++j){
                for(int d=0;d<=k;++d){
                    dp[i][j][d]=max(dp[i-1][j][d],dp[i][j-1][d]);
                    int cost=j-i;
                    if(d>=cost) dp[i][j][d]=max(dp[i][j][d],dp[i-1][j-1][d-cost]+a[j]-a[i]);
                }
            }
        }
        int ans,sum=0;
        ans=dp[L-1][n][k];
        for(int i=L;i<=R;++i) sum+=a[i];
        return sum-ans;
    }
}
namespace sub4{
    int dpL[N][N][N*N/4],dpR[N][N][N*N/4];
    int solve(){
        k=min(k,n*n/4);
        for(int i=0;i<=n+1;++i){
            for(int j=0;j<=n+1;++j){
                for(int d=0;d<=k;++d){
                    dpL[i][j][d]=0;
                    dpR[i][j][d]=0;
                }
            }
        }
        for(int i=1;i<L;++i){
            for(int j=L;j<=R;++j){
                for(int d=0;d<=k;++d){
                    dpL[i][j][d]=max(dpL[i-1][j][d],dpL[i][j-1][d]);
                    int cost=j-i;
                    if(d>=cost) dpL[i][j][d]=max(dpL[i][j][d],dpL[i-1][j-1][d-cost]+a[j]-a[i]);
                }
            }
        }
        for(int i=n;i>R;--i){
            for(int j=R;j>=L;--j){
                for(int d=0;d<=k;++d){
                    dpR[i][j][d]=max(dpR[i+1][j][d],dpR[i][j+1][d]);
                    int cost=i-j;
                    if(d>=cost) dpR[i][j][d]=max(dpR[i][j][d],dpR[i+1][j+1][d-cost]+a[j]-a[i]);
                }
            }
        }
        int ans=0,sum=0;
        for(int i=L-1;i<=R;++i){
            for(int d=0;d<=k;++d){
                ans=max(ans,dpL[L-1][i][d]+dpR[R+1][i+1][k-d]);
            }
        }
        for(int i=L;i<=R;++i) sum+=a[i];
//        cout<<sum<<" "<<ans<<endl;
        return sum-ans;
    }
}
int main(){
    ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
//    freopen("DEBT.inp","r",stdin);
//    freopen("DEBT.out","w",stdout);
    cin>>n>>L>>R>>k;
    for(int i=1;i<=n;++i) cin>>a[i];
//    if(n<=50 and R==n) cout<<sub2::solve();
//    else 
    cout<<sub4::solve();
    return 0;
}
/*
==================================+
INPUT:                            |
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------------------------------    |
==================================+
OUTPUT:                           |
------------------------------    |

------------------------------    |
==================================+
*/

• Subtask 1: 22 / 22
• Subtask 2: 33 / 33
• Subtask 3: 33 / 33
• Subtask 4: 22 / 22