QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#376064#6631. Maximum Bitwise OR369PaiTL 498ms66688kbC++142.3kb2024-04-03 20:14:592024-04-03 20:14:59

Judging History

你现在查看的是最新测评结果

  • [2024-04-03 20:14:59]
  • 评测
  • 测评结果:TL
  • 用时:498ms
  • 内存:66688kb
  • [2024-04-03 20:14:59]
  • 提交

answer

#include<bits/stdc++.h>
#define lc(p) ((p) * 2)
#define rc(p) ((p) * 2 + 1)
using namespace std;
typedef long long ll;
const int N = 1e5 + 5 , T = N * 4 , LG = 30;
int n , q , a[N];
struct Info
{
	int mx = 0 , s = 0 , ans = 0 , cnt[LG] = {}; 
	vector<int>v;
}tr[T];
ostream& operator << (ostream &os , const Info &a)
{
	os << a.mx << ' ' << a.s << ' ' << a.ans << "; ";
	for(int i = 0 ; (1ll << i) < a.mx ; i++)
		os << a.cnt[i] << " ";
	os << "; ";
	for(int x : a.v)os << x << " ";
	return os;
}
Info operator + (const Info &a , const Info &b)
{
	Info c; int s1 = 0 , s0 = 0;
	c.s = a.s | b.s , c.mx = a.mx | b.mx;
	c.ans = min(a.ans , b.ans);
	if(c.s == c.mx)c.ans = 0;
	for(int i = 0 ; i < LG ; i++)
	{
		c.cnt[i] = a.cnt[i] + b.cnt[i];
		if(!c.cnt[i])s0 |= 1 << i;
		if(c.cnt[i] == 1)s1 |= 1 << i;
	}
	auto chk = [&](int x)
	{
		if(!(x & s1))
		{
			if(c.ans <= 1)return ;
			for(int i = 0 ; i < LG ; i++)
			{
				int y = x ^ (x - (1 << i));
				if((s0 & y) == s0)
				{
					c.ans = min(c.ans , 1);
					break ;
				}
			}
		}
		else c.v.push_back(x);
	};
	for(int x : a.v)chk(x);
	for(int x : b.v)chk(x);
	// cerr << "Merge " << a << "\n+\n" << b << "\n";
	// cerr << "=\n" << c << "\n";
	return c;
}
void Pushup(int p){tr[p] = tr[lc(p)] + tr[rc(p)];}
void Build(int l , int r , int p)
{
	if(l == r)
	{
		tr[p].mx = (1ll << (64 - __builtin_clzll(a[l]))) - 1;
		tr[p].s = a[l] , tr[p].ans = 2;
		for(int i = 0 ; i < LG ; i++)
			tr[p].cnt[i] = (a[l] >> i) & 1;
		if(a[l])tr[p].v.push_back(a[l]);
		return ;
	}
	int mid = (l + r) >> 1;
	Build(l , mid , lc(p));
	Build(mid + 1 , r , rc(p));
	Pushup(p);
}
Info Query(int s , int e , int l = 1 , int r = n , int p = 1)
{
	if(s <= l && r <= e)return tr[p];
	int mid = (l + r) >> 1;
	if(s <= mid && mid < e)return Query(s , e , l , mid , lc(p)) + Query(s , e , mid + 1 , r , rc(p));
	else if(s <= mid)return Query(s , e , l , mid , lc(p));
	else return Query(s , e , mid + 1 , r , rc(p));
}
int Solve()
{
	cin >> n >> q;
	for(int i = 1 ; i <= n ; i++)cin >> a[i];
	Build(1 , n , 1);
	while(q--)
	{
		int l , r; cin >> l >> r;
		Info ans = Query(l , r);
		cout << ans.mx << ' ' << ans.ans << "\n";
	}
	return 0;
}
int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0) , cout.tie(0);
	int T; cin >> T;
	while(T--)Solve();
	return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 12ms
memory: 66108kb

input:

1
3 2
10 10 5
1 2
1 3

output:

15 2
15 0

result:

ok 4 number(s): "15 2 15 0"

Test #2:

score: 0
Accepted
time: 498ms
memory: 66688kb

input:

100000
1 1
924704060
1 1
1 1
149840457
1 1
1 1
515267304
1 1
1 1
635378394
1 1
1 1
416239424
1 1
1 1
960156404
1 1
1 1
431278082
1 1
1 1
629009153
1 1
1 1
140374311
1 1
1 1
245014761
1 1
1 1
445512399
1 1
1 1
43894730
1 1
1 1
129731646
1 1
1 1
711065534
1 1
1 1
322643984
1 1
1 1
482420443
1 1
1 1
20...

output:

1073741823 2
268435455 2
536870911 2
1073741823 2
536870911 2
1073741823 2
536870911 2
1073741823 2
268435455 2
268435455 2
536870911 2
67108863 2
134217727 2
1073741823 2
536870911 2
536870911 2
268435455 2
536870911 2
536870911 2
536870911 2
268435455 2
268435455 2
1073741823 2
16777215 2
10737418...

result:

ok 200000 numbers

Test #3:

score: -100
Time Limit Exceeded

input:

50000
2 2
924896435 917026400
1 2
1 2
2 2
322948517 499114106
1 2
2 2
2 2
152908571 242548777
1 1
1 2
2 2
636974385 763173214
1 2
1 1
2 2
164965132 862298613
1 1
1 2
2 2
315078033 401694789
1 2
1 2
2 2
961358343 969300127
2 2
1 2
2 2
500628228 28065329
1 2
1 2
2 2
862229381 863649944
1 2
2 2
2 2
541...

output:

1073741823 2
1073741823 2
536870911 2
536870911 2
268435455 2
268435455 2
1073741823 2
1073741823 2
268435455 2
1073741823 2
536870911 2
536870911 2
1073741823 2
1073741823 2
536870911 2
536870911 2
1073741823 2
1073741823 2
1073741823 2
268435455 2
536870911 2
536870911 2
1073741823 2
1073741823 2
...

result: