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>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<vpii> vvpii;
typedef vector<vpll> vvpll;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) (int)(x).size()
#define fi first
#define se second
template<class T> bool ckmin(T &a, const T &b) {return a > b ? a = b, 1 : 0;}
template<class T> bool ckmax(T &a, const T &b) {return a < b ? a = b, 1 : 0;}
namespace debug {
void __print(int x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for(auto z : x) cerr << (f++ ? "," : ""), __print(z); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if(sizeof...(v)) cerr << ", "; _print(v...);}
#ifdef ljuba
#define dbg(x...) cerr << "\e[91m" << "LINE(" << __LINE__ << ") -> " << "[" << #x << "] = ["; _print(x)
#else
#define dbg(x...)
#endif
}
using namespace debug;
const char nl = '\n';
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
/*
погледај сампле тестове, пре него што имплементираш (можда имаш погрешну идеју)
уместо да разматраш неке глупе случајеве на папиру, размишљај заправо о задатку
*/
const int LG = 43;
const ll INF = 1e18 + 12;
void solve() {
//nice problem, but different approach for different subtasks is ugly
int n, a, b;
cin >> n >> a >> b;
vll v(n);
for(auto &z : v) cin >> z;
vll pref(n+1);
for(int i = 0; i < n; ++i) {
pref[i+1] = pref[i] + v[i];
}
ll ans = (1LL<<LG) - 1;
auto moze = [&](ll x) {
if(a == 1) {
vll dp(n+1, INF);
dp[0] = 0;
for(int i = 1; i <= n; ++i) {
for(int j = i-1; j >= 0; --j) {
ll suma = pref[i] - pref[j];
if((suma | x) != x) continue;
//1101110
//1010000
// .
ckmin(dp[i], dp[j] + 1);
}
}
return dp[n] <= b;
}
vector<vector<bool>> dp(n+1, vector<bool>(n+1, false));
dp[0][0] = true;
for(int i = 1; i <= n; ++i) {
for(int j = i-1; j >= 0; --j) {
for(int k = 0; k <= n; ++k) {
ll suma = pref[i] - pref[j];
if((suma | x) != x) continue;
if(dp[j][k]) dp[i][k+1] = true;
// dp[i][k+1] = dp[j][k];
}
}
}
for(int i = a; i <= b; ++i) {
if(dp[n][i]) return true;
}
return false;
};
for(int i = LG-1; i >= 0; --i) {
if(moze(ans-(1LL<<i))) {
ans -= (1LL<<i);
}
}
cout << ans << nl;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int testCases = 1;
//cin >> testCases;
while(testCases--)
solve();
}
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