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>
#ifdef OP_DEBUG
#include <algo/debug.h>
#else
#define debug(...) 26
#endif
using namespace std;
#define int long long // maybe tle?
#define sz(v) (int)(v).size()
#define all(v) (v).begin(), (v).end()
#define TcT template <class T
const int MOD = (int)1e9 + 7, INF = (int)2e9 + 9, INF64 = (int)4e18 + 18;
TcT> bool minimize(T &lhs, const T &rhs) { return rhs < lhs ? lhs = rhs, 1 : 0; }
TcT> bool maximize(T &lhs, const T &rhs) { return rhs > lhs ? lhs = rhs, 1 : 0; }
TcT, class S> istream &operator >> (istream &scanf, pair <T, S> &u) { return scanf >> u.first >> u.second; }
TcT, class S> ostream &operator << (ostream &print, pair <T, S> &u) { return print << u.first << ' ' << u.second; }
void solve();
int32_t main() {
cin.tie(nullptr), cout.tie(nullptr) -> sync_with_stdio(0);
int testcases = 1;
#define TC 0
if (TC) { cin >> testcases; } for (; testcases--;) { solve(); }
}
/* [Pham Hung Anh - 11I - Tran Hung Dao High School for Gifted Student] */
const int N = 2002;
int n, a, b;
int y[N];
namespace sub1 {
int ans = INF;
void Try(int k, int parts, int sum, int sum_or) {
if (k > n) {
if (a <= parts && parts <= b)
minimize(ans, sum_or | sum);
return;
}
Try(k + 1, parts, sum + y[k], sum_or);
Try(k + 1, parts + 1, y[k], sum_or | sum);
}
void solve() {
Try(1, 1, 0, 0);
cout << ans << '\n';
}
}
namespace sub2 {
// max ans = 500?
bool dp[55][22][505]; // xét đến i, đoạn thứ parts, tổng or là sum_or
void solve() {
dp[0][0][0] = 1;
for (int i = 1; i <= n; ++i)
for (int parts = 0; parts <= b; ++parts)
for (int sum_or = 0; sum_or <= 500; ++sum_or) {
int sum = 0;
for (int j = i; j >= 1; --j)
dp[i][parts + 1][sum_or | (sum += y[j])] |= dp[j - 1][parts][sum_or];
}
int ans = INF;
for (int parts = a; parts <= b; ++parts)
for (int sum_or = 0; sum_or <= 500; ++sum_or)
if (dp[n][parts][sum_or])
minimize(ans, sum_or);
cout << ans << '\n';
}
}
void solve() {
cin >> n >> a >> b;
for (int i = 1; i <= n; ++i)
cin >> y[i];
if (n <= 20)
sub1 :: solve();
else if (n <= 50)
sub2 :: solve();
}
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