제출 #41432

#제출 시각아이디문제언어결과실행 시간메모리
41432cmasterBali Sculptures (APIO15_sculpture)C++14
100 / 100
256 ms876 KiB
#include <bits/stdc++.h>
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>*/
 
#define pb push_back
#define mp make_pair
#define sz(s) ((int)(s.size()))
#define all(s) s.begin(), s.end()
#define rep(i, a, n) for (int i = a; i <= n; ++i)
#define per(i, n, a) for (int i = n; i >= a; --i)
#define onlycin ios_base::sync_with_stdio(false); cin.tie(0) 
#define F first
#define S second
using namespace std;
// using namespace __gnu_pbds;
typedef long long ll;
typedef unsigned long long ull;
/*typedef tree<
pair < int, int >,
null_type,
less< pair < int, int > >,
rb_tree_tag,
tree_order_statistics_node_update>
ordered_set;*/
// find_by_order() order_of_key()
const int MAXN = (int)5e5+228;
const char nxtl = '\n';
const int mod = (int)1e9+7;
const double eps = (double)1e-7;
template<typename T> inline bool updmin(T &a, const T &b) {return a > b ? a = b, 1 : 0;}
template<typename T> inline bool updmax(T &a, const T &b) {return a < b ? a = b, 1 : 0;}
 
int n, l, r, a[MAXN];
int dp[MAXN], dp2[MAXN];
 
inline bool check(const ll &lim, const ll &mid, const ll &MOD, const ll &tmp) {
	dp[0] = 0;
	dp2[0] = 0;
	rep(i, 1, n) {
		ll cur = 0ll;
		dp[i] = n+n;
		dp2[i] = -(n+n);
		per(j, i, 1) {
			cur += a[j];
			if(cur >= lim) break;
			if((cur & tmp) == 0ll && cur % MOD < mid) {
				updmin(dp[i], dp[j-1]+1);
				updmax(dp2[i], dp2[j-1]+1);
			}
		}
	}
	return dp[n] <= r && dp2[n] >= l;
}
map < pair < ll, pair < ll, ll > >, ll > dd;
ll rec(int i, ll cost, int cnt) {
	if(i == n+1) {
		if(l <= cnt && cnt <= r) return cost;
		return (ll)1e15;
	}
	if(dd.count(mp(i, mp(cost, cnt)))) return dd[mp(i, mp(cost, cnt))];
	ll ans = (ll)1e15;
	ll sum = 0;
	rep(j, i, n) {
		sum += a[j];
		ans = min(ans, rec(j+1, cost | sum, cnt+1));
	}
	return dd[mp(i, mp(cost, cnt))] = ans;
}
 
int main() {
	#ifdef accepted
		freopen(".in", "r", stdin);
		freopen(".out", "w", stdout);
	#endif
	onlycin;
	cin >> n >> l >> r;
	rep(i, 1, n) cin >> a[i];
	if(l > 1 && n <= 50) {
		cout << rec(1, 0, 0) << nxtl;
		return 0;
	}
	int last = 42, fr = 42;
	ll res = 0ll;
	ll just = 0ll;
	per(bit, 40, 0) {
		ll lim = (1ll << fr*1ll);
		if(last == 42) lim = (1ll << bit);
		if(!check(lim, (1ll<<bit*1ll), (1ll << last*1ll), just)) {
			if(fr == 42) fr = bit+1;
			res |= (1ll << bit*1ll);
			last = bit;
		} else just |= (1ll << bit*1ll);
	}
	cout << res << nxtl;
	
	return 0;
}
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