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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#300900#7967. 二进制defyers#ML 4ms60764kbC++203.2kb2024-01-08 23:50:092024-01-08 23:50:10

Judging History

你现在查看的是最新测评结果

  • [2024-01-08 23:50:10]
  • 评测
  • 测评结果:ML
  • 用时:4ms
  • 内存:60764kb
  • [2024-01-08 23:50:09]
  • 提交

answer

#include "bits/stdc++.h"
using namespace std;

#pragma GCC optimize("Ofast")
#pragma GCC target("avx2")

using ll = long long;
using LL = long long;

const int N = 1 << 20;

struct BIT{
	int bit[N];
	void add(int p, int x){ for(int i = p; i < N; i += i & -i) bit[i] += x; }
	int qry(int p){ int ret = 0; for(int i = p; i; i -= i & -i) ret += bit[i]; return ret; }
	int bs(int v){ int ret = 0, s = 0; for(int j = 19; j >= 0; j--) if(ret + (1 << j) < N && s + bit[ret + (1 << j)] < v) ret += (1 << j), s += bit[ret]; return ret + 1; }
} bit;

int n; 
bool a[N];
int l[N], r[N];
set<int> S[N];

int L[20], R[20];
int cur_len;
void rem(int x){
	int sl = 0, sr = 0;
	for(int y = l[x], tr = 0; y != 0 && tr < 20; tr++){
		L[tr] = y; y = l[y]; sl++;
	}
	for(int y = r[x], tr = 0; y != n + 1 && tr < 20; tr++){
		R[tr] = y; y = r[y]; sr++;
	}

	for(int i = sl - 1; i >= 0; i--){
		if(!a[L[i]]) continue;
		int v = 1;
		for(int j = i - 1; j >= 0; j--){
			v = v * 2 + a[L[j]];
		}
		// NOT choose middle
		int u = v;
		for(int j = 0; j < sr; j++){
			u = u * 2 + a[R[j]];
			// if(u >= N) break;
			// if(u > cur_len)
			// 	S[u].insert(L[i]);
		}
		// DOES choose middle
		u = v * 2 + a[x];
		if(u < N && u > cur_len) S[u].erase(L[i]);
		// cout << "R " << u << ' ' << L[i] << endl;
		for(int j = 0; j < sr; j++){
			u = u * 2 + a[R[j]];
			if(u >= N) break;
			// cout << "ADD " << u << ' ' << L[i] << endl;
			if(u > cur_len)
				S[u].erase(L[i]);
		}
	}

	if(a[x]){
		int u = a[x];
		if(u > cur_len) S[u].erase(x);
		for(int j = 0; j < sr; j++){
			u = u * 2 + a[R[j]];
			if(u >= N) break;
			if(u > cur_len) S[u].erase(x);
		}
	}
	for(int i = sl - 1; i >= 0; i--){
		if(!a[L[i]]) continue;
		int v = 1;
		for(int j = i - 1; j >= 0; j--){
			v = v * 2 + a[L[j]];
		}
		// NOT choose middle
		int u = v;
		for(int j = 0; j < sr; j++){
			u = u * 2 + a[R[j]];
			if(u >= N) break;
			if(u > cur_len)
				S[u].insert(L[i]);
		}
	}
	bit.add(x, 1);
	l[r[x]] = l[x];
	r[l[x]] = r[x];
}

int len(int l){
	int x = 0;
	while(l){
		x++;
		l /= 2;
	}
	return x;
}

void solve(int TC) {
	cin >> n;
	for(int i = 0; i <= n + 1; i++){
		l[i] = i - 1;
		r[i] = i + 1;
	}
	l[0] = 0, r[n + 1] = n + 1;
	string s; cin >> s;
	for(int i = 1; i <= n; i++){
		a[i] = s[i - 1] == '1';
	}

	for(int i = 1; i <= n; i++){
		if(a[i] == 0) continue;
		int v = 0;
		for(int j = i; j < i + 20 && j <= n; j++){
			v = v * 2 + a[j];
			S[v].insert(i);
		}
	}

	// for(int l = 1; l <= n; l++){
	// 	for(auto u : S[l]) cout << u << ' '; cout << '\n';
	// }

	for(int l = 1; l <= n; l++){
		cur_len = l;
		int cnt = S[l].size();
		if(cnt == 0){
			cout << -1 << ' ' << 0 << '\n';
		}else{
			int idx = 0;
			assert(cnt > 0);
			// cout << "S T" << endl;
			// for(auto x : S[l]) cout << x << ' '; cout << '\n';
			// for(auto x : T[l]) cout << x << ' '; cout << '\n';
			int pos = *S[l].begin();
			cout << pos - bit.qry(pos - 1) << ' ' << cnt << '\n';
			for(int j = 0; j < len(l); j++){
				rem(pos); pos = r[pos];
			}
		}
		S[l].clear();
	}
}

int32_t main() {
	cin.tie(0)->sync_with_stdio(0);
	cout << fixed << setprecision(10);

	int t = 1;
	// cin >> t;

	for (int i = 1; i <= t; i++) {
		solve(i);
	}
}

详细

Test #1:

score: 100
Accepted
time: 4ms
memory: 60764kb

input:

20
01001101101101110010

output:

2 11
5 5
4 5
11 1
4 2
7 1
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0

result:

ok 20 lines

Test #2:

score: -100
Memory Limit Exceeded

input:

1000000
1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111...

output:

1 1000000
-1 0
1 999998
-1 0
-1 0
-1 0
1 999995
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
1 999991
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
1 999986
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0
-1 0...

result: