QOJ.ac

QOJ

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
#851758#9575. $P \oplus Q = R$BUET_ALUCHASHI#WA 1ms3748kbC++204.2kb2025-01-10 23:32:502025-01-10 23:32:50

Judging History

你现在查看的是最新测评结果

  • [2025-01-10 23:32:50]
  • 评测
  • 测评结果：WA
  • 用时：1ms
  • 内存：3748kb
查看
  • [2025-01-10 23:32:50]
  • 提交

answer

#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#include <time.h>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <cstring>

using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <iostream>

using namespace __gnu_pbds;
using namespace std;
template <class T>
using Tree =
    tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// to erase in multiset-> less_equal<T> and 
// s.erase(s.find_by_order(s.order_of_key(x)))
// lower_bound(x)=>(cannot use the stl lower_bound function)
// ll idx = s.order_of_key(x)
// if(idx == s.size()) -> no lower_bound
// else lb = *s.find_by_order(idx) // as 0-indexing
// idx-1 will give highest value which is strictly less than x
// for upper_bound->do the same with (x+1)

typedef long long ll;
typedef long double ld;
typedef pair<int,int> p32;
typedef pair<ll,ll> p64;
typedef tuple<ll, ll, ll> t64;
typedef vector<t64> vt64;
typedef vector<vt64> vvt64;
typedef pair<double,double> pdd;
typedef vector<ll> v64;
typedef vector<int> v32;
typedef vector<vector<int> > vv32;
typedef vector<vector<ll> > vv64;
typedef vector<vector<p64> > vvp64;
typedef vector<p64> vp64;
typedef vector<p32> vp32;
typedef vector<vector<p32> > vvp32;
typedef vector<bool> vb;
ll mod =  1e9+7, MOD = 998244353;
double eps = 1e-12;
#define FOR(s, e, i) for(ll i = s; i <= e; i++)
#define ROF(s ,e, i) for(ll i = s; i >= e; i--)
#define F0R(i, e) for(ll i = 0; i < (e); i++)
#define trav(e, a) for(auto &e : a)
#define coutAll(A) for(auto asdafas : A) cout <<  asdafas << " "; cout << "\n";
#define foutAll(A) for(auto asdafas : A) fout <<  asdafas << " "; cout << "\n";
#define cinAll(A) for(auto &asdafas : A) cin >>  asdafas;
#define finAll(A) for(auto &asdafas : A) fin >>  asdafas;
#define minpq priority_queue<ll, v64, greater<ll>>
#define maxpq priority_queue<ll> 
#define ln "\n"
#define dbg(x) cout<<#x<<" = "<<x<<ln
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define fi first
#define se second
ll inf = LLONG_MAX;
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((ll)(x).size())
#define yes cout<<"yes\n"
#define no cout<<"no\n"
#define Yes cout<<"Yes\n"
#define No cout<<"No\n"
#define YES cout<<"YES\n"
#define NO cout<<"NO\n"
using namespace std;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef pair<ll, ll> pll;
typedef pair<ll, ll> pii;
#define MAXN 100000000

void solve(int it)
{
    // int n = it;
    // // cin >> n;
    // cout << it << "\n";
    // v64 A(n);
    // iota(all(A), 0);
    // v64 B = A;
    // do{
    //     set<ll>s;
    //     FOR(0, n - 1, i){
    //         s.insert(A[i]^B[i]);
    //     }
    //     if(sz(s)==n && *s.begin()==0 && *s.rbegin()==n-1){
    //         // coutAll(A);
    //         coutAll(B);
    //         // FOR(0, n - 1, i) cout << (A[i]^B[i]) << " ";
    //         // cout << '\n';
    //         cout << '\n';
    //     }
    // }while(next_permutation(all(B)));
    int n;
    cin >> n;
    if(n<=2){
        Yes;
        F0R(j, 2){
            FOR(0, n-1, i) cout << i << " ";
            cout << '\n';
        }
        return;
    }
    int bit = log2(n);
    if((1<<bit) != n){
        cout << "No\n";
        return;
    }
    v32 A(n);
    iota(all(A), 0);
    v64 B;
    FOR(0, n-1, i){
        B.pb(i);
        ++i;
    }
    FOR(1, n-3, i){
        B.pb(i+2);
        B.pb(i);
        i+=3;
    }
    Yes;
    coutAll(A);
    coutAll(B);
    // set<ll>s;
    // FOR(0, n - 1, i) {
    //     cout << (A[i]^B[i]) << " ";
    //     s.insert(A[i]^B[i]);
    // }
    // cout << "\n";
    // if(sz(s) == n && *s.rbegin() == n-1){
    //     Yes;
    // }
    // else{
    //     No;
    // }
}


int main()
{
    // freopen("text.txt", "w", stdout);
    fast_cin();    
    ll t = 1;
    cin >> t;
    for(int it=1; it<=t; it++)
    {
        //cout << "Case " << it << ": ";
        solve(it);
    }
    return 0;
}
源文件, 语言: C++20

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 3708kb

input:

2
3
4

output:

No
Yes
0 1 2 3 
0 2 3 1

result:

ok Correct. (2 test cases)

Test #2:

score: -100
Wrong Answer
time: 1ms
memory: 3748kb

input:

1999
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101...

output:

Yes
0 
0 
Yes
0 1 
0 1 
No
Yes
0 1 2 3 
0 2 3 1 
No
No
No
Yes
0 1 2 3 4 5 6 7 
0 2 4 6 3 1 7 5 
No
No
No
No
No
No
No
Yes
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 
0 2 4 6 8 10 12 14 3 1 7 5 11 9 15 13 
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
Yes
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20...

result:

wrong answer Testcase 2, n = 2, Participant's solution is incorrect. cntr[ 0] = 2 (test case 2)
English 繁體中文 简体中文

服务器时间: 2025-02-01 23:47:19 | 开源项目