Submission #610541

#	Time	Username	Problem	Language	Result	Execution time	Memory
610541		nohaxjustsoflo	Bali Sculptures (APIO15_sculpture)	C++17	100 / 100	180 ms	16144 KiB

sculpture

#include <bits/stdc++.h>
#include <iostream>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<ll,null_type,less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> order_set;
mt19937 mt_rand(chrono::high_resolution_clock::now().time_since_epoch().count());
//uniform_int_distribution<int> gen; ///(min, max)
//int random() {return gen(mt_rand);}
const int mxN=2001;
const int mod=998244353;
const int mxlogN=40;
const int mxK=26;
const int inf=1e9;
const int K=600;
int dp[mxN][mxN], a[mxN], n, A, B;
int vis[mxN];
ll check(int i, int k, ll msk, bool b)
{
    if(!b)
    {
        if(i==n) return k>=A&&k<=B;
        if(~dp[i][k]) return dp[i][k];
        ll sum=0;
        for(int j=i+1; j<=n; j++)
        {
            sum+=a[j-1];
            if((msk&sum)==sum) if(check(j,k+1,msk,0)) return dp[i][k]=1;
        }
        return dp[i][k]=0;
    }
    else
    {
        queue<int> q; q.push(0);
        for(int i=0; i<=n; i++) vis[i]=0;
        while(q.size())
        {
            int i=q.front(); q.pop();
            ll sum=0;
            for(int j=i+1; j<=n; j++)
            {
                sum+=a[j-1];
                if((msk&sum)==sum)
                {
                    if(!vis[j])
                    {
                        vis[j]=vis[i]+1;
                        q.push(j);
                    }
                }
            }
        }
        return vis[n]&&vis[n]<=B;
    }
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    for(int i=0; i<mxN; i++) for(int j=0; j<mxN; j++) dp[i][j]=-1;

    cin >> n >> A >> B;
    for(int i=0; i<n; i++) cin >> a[i];
    ll msk=(1LL<<(mxlogN+1))-1;
    for(int j=0; j<=mxlogN; j++)
    {
        msk^=check(0,0,msk^1LL<<(mxlogN-j),n>100)<<(mxlogN-j);
        for(int i=0; i<mxN; i++) for(int j=0; j<mxN; j++) dp[i][j]=-1;
    }
    cout << msk << "\n";
}
/*
6 2 2
8 1 2 1 5 4
*/

• Subtask 1: 9 / 9
• Subtask 2: 16 / 16
• Subtask 3: 21 / 21
• Subtask 4: 25 / 25
• Subtask 5: 29 / 29