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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
#122755#6631. Maximum Bitwise ORwnmrmr#WA 875ms3600kbC++234.5kb2023-07-11 01:49:462023-07-11 01:49:50

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-07-11 01:49:50]
  • 评测
  • 测评结果:WA
  • 用时:875ms
  • 内存:3600kb
  • [2023-07-11 01:49:46]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
 
#define all(x) x.begin(), x.end()
 
void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T){ cerr << ' ' << H; dbg_out(T...); }
#define dbg(...) //cerr<<"(" << #__VA_ARGS__<<"):" , dbg_out(__VA_ARGS__) , cerr << endl

#define pb push_back
#define int long long

int n, q;

void solve () {
    cin >> n >> q;
    vector<int> a (n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];

    vector last (n + 1, vector<int> (30));
    vector sum_bit (n + 1, vector<int> (30));
    vector freq (n + 1, vector (30, vector<int> (30)));
    vector acum (n + 1, vector (30, vector<int> (30)));
    for (int i = 1; i <= n; i++) {
        last[i] = last[i - 1];
        sum_bit[i] = sum_bit[i - 1];
        acum[i] = acum[i - 1];

        // vamos adicionar o numero
        for (int l = 0; l < 30; l++) {
            // atualizamos o ultimo
            if ((1 << l) & a[i]) last[i][l] = i, sum_bit[i][l]++;

            // atualizamos freq
            freq[i][l][l]++;
            for (int r = l + 1; r < 30; r++) {
                if ((1 << r) & a[i]) break;
                else freq[i][l][r]++;
            }
        }

        for (int l = 0; l < 30; l++) 
            for (int r = l; r < 30; r++)
                acum[i][l][r] += freq[i][l][r];
    }

    auto get_f = [&] (int l, int r) {
        vector f (30, vector<int> (30));
        for (int i = 0; i < 30; i++)
            for (int j = i; j < 30; j++)
                f[i][j] = acum[r][i][j] - acum[l - 1][i][j];
        return f;
    };
    auto get_sum = [&] (int l, int r) {
        vector<int> sum (30);
        for (int i = 0; i < 30; i++) sum[i] = sum_bit[r][i] - sum_bit[l - 1][i];
        return sum;
    };

    while (q--) {
        int l, r; cin >> l >> r;

        auto f = get_f (l, r);
        auto sum = get_sum (l, r);

        int mx = 29;
        while (mx >= 0 && !sum[mx]) mx--;
        // dbg (mx);
        cout << (1 << (mx + 1)) - 1 << " ";
        if (mx < 0) {
            // se o or eh 0
            cout << "0\n";
            continue;  
        }
        bool ok = true;
        for (int j = 0; j <= mx; j++) if (!sum[j]) ok = false;
        if (ok) {
            cout << "0\n";
            continue;
        }

        // senao, temos que ver se a resposta eh 1
        // vamos fazer operacao com cada cara 1, e depois vamos ver 

        vector<int> esp;
        for (int j = 0; j < 30; j++) if (sum[j] == 1) {
            // se esse cara eh unico
            // dbg (last[r][j]);
            esp.pb (last[r][j]);
        }
        // dbg (esp.size ());
        if (esp.size ()) {
            sort (all (esp));
            esp.resize (unique (all (esp)) - esp.begin ());
        }

        ok = false;
        for (auto id : esp) {
            // retiramos e vermos se eh possivel
            for (int i = 0; i < 30; i++) {
                if ((1 << i) & a[id]) sum[i]--;
                for (int j = i; j < 30; j++) {
                    f[i][j] -= freq[id][i][j];
                }
            }

            // temos que ver se ainda existe um cara que preenche
            int esq = 30, dir = -1;
            for (int j = 0; j <= mx; j++) if (!sum[j]) {
                esq = min (esq, j);
                dir = max (dir, j);
            }
            // dbg (esq, dir); 

            if (freq[id][esq][dir]) {
                ok = true;
            }

            for (int i = 0; i < 30; i++) {
                if ((1 << i) & a[id]) sum[i]++;
                for (int j = i; j < 30; j++) {
                    f[i][j] += freq[id][i][j];
                }
            }
        }

        if (ok) {
            cout << "1\n";
            continue;
        }

        for (auto id : esp) {
            // retiramos e vermos se eh possivel
            for (int i = 0; i < 30; i++) {
                if ((1 << i) & a[id]) sum[i]--;
                for (int j = i; j < 30; j++) {
                    f[i][j] -= freq[id][i][j];
                }
            }
        }

        int esq = 30, dir = -1;
        for (int j = 0; j <= mx; j++) if (!sum[j]) {
            esq = min (esq, j);
            dir = max (dir, j);
        }

        if (f[esq][dir]) {
            ok = true;
        }

        if (ok) {
            cout << "1\n";
            continue;
        }

        cout << "2\n";
    }
}

signed main () {
    ios::sync_with_stdio(0);cin.tie(0);
    int t; cin >> t;
    while (t--)
        solve ();
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 3500kb

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: 875ms
memory: 3460kb

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: 0
Accepted
time: 747ms
memory: 3544kb

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:

ok 200000 numbers

Test #4:

score: -100
Wrong Answer
time: 755ms
memory: 3600kb

input:

33333
3 3
925088809 339284112 289540728
3 3
1 3
1 1
3 3
422399522 892365243 216341776
3 3
3 3
1 2
3 3
668932010 837523227 840095874
1 3
1 3
3 3
3 3
731584574 357877180 359063739
1 1
1 1
3 3
3 3
463358343 833924976 847087403
2 3
3 3
1 2
3 3
377154649 772000701 656357011
2 3
1 2
2 3
3 3
977492169 5540...

output:

536870911 2
1073741823 2
1073741823 2
268435455 2
268435455 2
1073741823 2
1073741823 2
1073741823 2
1073741823 2
1073741823 2
1073741823 2
536870911 2
1073741823 2
1073741823 2
1073741823 2
1073741823 2
1073741823 2
1073741823 2
1073741823 2
1073741823 2
1073741823 2
67108863 2
1073741823 2
1073741...

result:

wrong answer 3718th numbers differ - expected: '2', found: '1'