이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define FAST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);cerr.tie(0)
#define mp make_pair
#define xx first
#define yy second
#define pb push_back
#define pf push_front
#define popb pop_back
#define popf pop_front
#define all(x) x.begin(),x.end()
#define ff(i,a,b) for (int i = a; i < b; i++)
#define fff(i,a,b) for (int i = a; i <= b; i++)
#define bff(i,a,b) for (int i = b-1; i >= a; i--)
#define bfff(i,a,b) for (int i = b; i >= a; i--)
using namespace std;
long double typedef ld;
unsigned int typedef ui;
long long int typedef li;
pair<int,int> typedef pii;
pair<li,li> typedef pli;
pair<ld,ld> typedef pld;
vector<vector<int>> typedef graph;
unsigned long long int typedef ull;
//const int mod = 998244353;
const int mod = 1000000007;
//Note to self: Check for overflow
li a[2006];
li pre[2006];
bool mz[103][103];
li brute(int n, int A, int B)
{
li ans=0;
li include=0;
bff(b,0,50)
{
li cur=include|(1ll<<b);
fff(k,0,n) fff(i,1,n) mz[k][i]=0;
mz[0][n+1]=1;
fff(k,1,n) fff(i,1,n) fff(j,i,n)
if (!((pre[j]-pre[i-1])&cur)) mz[k][i]|=mz[k-1][j+1];
bool los=true;
fff(k,A,B) if (mz[k][1]) los=false;
if (los) ans|=(1ll<<b);
else include=cur;
}
return ans;
}
int dp[2006];
int main()
{
FAST;
int n,A,B; cin>>n>>A>>B;
fff(i,1,n) cin>>a[i];
fff(i,1,n) pre[i]=pre[i-1]+a[i];
if (n<=100) return cout<<brute(n,A,B),0;
int k=B;
li ans=0;
li include=0;
bff(b,0,50)
{
li cur=include|(1ll<<b);
fff(i,1,n) dp[i]=mod;
dp[n+1]=0;
bfff(i,1,n)
{
li tmp=0;
fff(j,i,n)
{
tmp+=a[j];
if (!(tmp&cur)) dp[i]=min(dp[i],dp[j+1]+1);
}
}
if (dp[1]>k) ans|=(1ll<<b);
else include=cur;
}
cout<<ans<<"\n";
}
//Note to self: Check for overflow
/*
6 1 3
8 1 2 1 5 4
10 1 10
1 1 1 1 1 1 1 1 1 1
*/
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