이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ld,ld> pdd;
#define ff(i,a,b) for(int i = a; i <= b; i++)
#define fb(i,b,a) for(int i = b; i >= a; i--)
#define trav(a,x) for (auto& a : x)
#define sz(a) (int)(a).size()
#define pb push_back
#define fi first
#define se second
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k) print the k-th smallest number in os(0-based)
const int mod = 1000000007;
const int inf = 1e9 + 5;
const int mxN = 2005;
int n, A, B;
ll niz[mxN];
bool dp[mxN][mxN];
bool check(ll msk){
// da li imam resenje tako da je svaki bit iz mask
if(A == 1){
bool ok = 0;
ff(i,1,n)if((niz[i] & msk) != niz[i])ok = 1;
if(ok == 0)return 1;
}
ff(i,0,n)ff(j,0,B)dp[i][j] = 0;
dp[0][0] = 1;
ff(i,1,n){
ff(j,1,B){
ll X = 0;
fb(l,i,1){
X += niz[l];
if((X & msk) == X){
dp[i][j] |= dp[l - 1][j - 1];
}
}
}
}
ff(j,A,B)if(dp[n][j] == 1)return 1;
return 0;
}
int main() {
cin.tie(0)->sync_with_stdio(0);
cin >> n >> A >> B;
ff(i,1,n)cin >> niz[i];
// znaci ja hocu da or bude minimalan
ll rez = (1ll << 41) - 1;
fb(i,40,0){
if(check(rez ^ (1ll << i))){
rez ^= (1ll << i);
}
}
cout << rez << '\n';
return 0;
}
/*
6 1 3
8 1 2 1 5 4
// probati bojenje sahovski
*/
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