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322 lines
9.1 KiB
Python
Executable file
322 lines
9.1 KiB
Python
Executable file
"""#
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### 谜题描述
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This is an interactive problem!
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Nastia has a hidden permutation p of length n consisting of integers from 1 to n. You, for some reason, want to figure out the permutation. To do that, you can give her an integer t (1 ≤ t ≤ 2), two different indices i and j (1 ≤ i, j ≤ n, i ≠ j), and an integer x (1 ≤ x ≤ n - 1).
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Depending on t, she will answer:
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* t = 1: max{(min{(x, p_i)}, min{(x + 1, p_j)})};
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* t = 2: min{(max{(x, p_i)}, max{(x + 1, p_j)})}.
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You can ask Nastia at most ⌊ \frac {3 ⋅ n} { 2} ⌋ + 30 times. It is guaranteed that she will not change her permutation depending on your queries. Can you guess the permutation?
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Input
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The input consists of several test cases. In the beginning, you receive the integer T (1 ≤ T ≤ 10 000) — the number of test cases.
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At the beginning of each test case, you receive an integer n (3 ≤ n ≤ 10^4) — the length of the permutation p.
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It's guaranteed that the permutation is fixed beforehand and that the sum of n in one test doesn't exceed 2 ⋅ 10^4.
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Interaction
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To ask a question, print \"? t i j x\" (t = 1 or t = 2, 1 ≤ i, j ≤ n, i ≠ j, 1 ≤ x ≤ n - 1) Then, you should read the answer.
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If we answer with −1 instead of a valid answer, that means you exceeded the number of queries or made an invalid query. Exit immediately after receiving −1 and you will see the Wrong Answer verdict. Otherwise, you can get an arbitrary verdict because your solution will continue to read from a closed stream.
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To print the answer, print \"! p_1 p_2 … p_{n} (without quotes). Note that answering doesn't count as one of the ⌊ \frac {3 ⋅ n} {2} ⌋ + 30 queries.
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After printing a query or printing the answer, do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
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* fflush(stdout) or cout.flush() in C++;
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* System.out.flush() in Java;
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* flush(output) in Pascal;
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* stdout.flush() in Python;
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* See the documentation for other languages.
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Hacks
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To hack the solution, use the following test format.
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The first line should contain a single integer T (1 ≤ T ≤ 10 000) — the number of test cases.
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For each test case in the first line print a single integer n (3 ≤ n ≤ 10^4) — the length of the hidden permutation p.
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In the second line print n space-separated integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n), where p is permutation.
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Note that the sum of n over all test cases should not exceed 2 ⋅ 10^4.
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Example
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Input
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2
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4
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3
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2
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5
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3
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Output
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? 2 4 1 3
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? 1 2 4 2
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! 3 1 4 2
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? 2 3 4 2
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! 2 5 3 4 1
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Note
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Consider the first test case.
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The hidden permutation is [3, 1, 4, 2].
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We print: \"? 2 4 1 3\" and get back min{(max{(3, p_4}), max{(4, p_1)})} = 3.
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We print: \"? 1 2 4 2\" and get back max{(min{(2, p_2)}, min{(3, p_4)})} = 2.
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Consider the second test case.
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The hidden permutation is [2, 5, 3, 4, 1].
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We print: \"? 2 3 4 2\" and get back min{(max{(2, p_3}), max{(3, p_4)})} = 3.
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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#include \"bits/stdc++.h\"
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#pragma GCC optimize(\"Ofast\")
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#pragma GCC target(\"avx,avx2,fma\")
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// #pragma GCC optimization (\"O3\")
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// #pragma GCC optimization (\"unroll-loops\")
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// #include <ext/pb_ds/assoc_container.hpp>
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// #include <ext/pb_ds/tree_policy.hpp>
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// using namespace __gnu_pbds;
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//#include <boost/multiprecision/cpp_int.hpp>
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//using namespace boost::multiprecision;
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using namespace std;
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#define all(c) (c).begin(),(c).end()
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// #define endl \"\n\"
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#define ff first
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#define ss second
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#define allr(c) (c).rbegin(),(c).rend()
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#define ifr(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
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#define pof pop_front
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#define pob pop_back
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#define pb emplace_back
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#define pf emplace_front
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#define fstm(m,n,r) m.reserve(n);m.max_load_factor(r)
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#define mp make_pair
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#define mt make_tuple
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#define inf LLONG_MAX
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#define os tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
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//order_of_key (k) : Number of items strictly smaller than k .
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//find_by_order(k) : K-th element in a set (counting from zero).
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const double PI = acos(-1);
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typedef complex<double> cd;
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typedef long long ll;
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ll gcd(){return 0ll;} template<typename T,typename... Args> T gcd(T a,Args... args) { return __gcd(a,(__typeof(a))gcd(args...)); }
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typedef map<ll,ll> mll;
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typedef map<string,ll> msll;
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typedef unordered_map<ll,ll> umap;
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typedef vector<ll> vll;
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typedef pair<ll,ll> pll;
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typedef long double ld;
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#define mod 1000000007
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const int N = 1e5 + 2;
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int ask(int t, int i, int j, int x) {
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int r;
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if(t == 1) t = 2; else t = 1;
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cout<<\"? \"<<t<<\" \"<<i+1<<\" \"<<j+1<<\" \"<<x<<endl;
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cin>>r;
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return r;
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}
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void answer(vector<int> &a) {
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cout<<\"! \";
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for(auto &it: a)
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cout<<it<<\" \";
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cout<<endl;
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}
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void Solve() {
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int n;
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cin>>n;
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int x, y;
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vector<int> a(n,0);
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for(int i=0; i+1<n; i+=2) {
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x = ask(1, i, i+1, 1);
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y = ask(2, i, i+1, n-1);
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if(x == 2 && ask(2, i, i+1, 1)==1) {
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a[i+1] = 1;
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if(y == n-1 && ask(1, i, i+1, n-1)==n)
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a[i] = n;
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else
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a[i] = y;
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continue;
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}
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if(y == n-1 && ask(1, i, i+1, n-1) == n) {
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a[i] = n;
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a[i+1] = x;
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continue;
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}
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if(ask(1, i+1, i, x) == x+1) {
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a[i] = x;
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a[i+1] = y;
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} else {
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a[i] = y;
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a[i+1] = x;
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}
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}
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if(n&1) {
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vector<bool> vis(n+1, 0);
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for(int i=0; i<n-1; i++)
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vis[a[i]] = 1;
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for(int i=1; i<n+1; i++) if(!vis[i]) {
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a[n-1] = i;
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break;
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}
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}
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answer(a);
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}
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int main() {
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ios_base::sync_with_stdio(false);
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cin.tie(NULL);
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cout.tie(NULL);
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// #ifndef ONLINE_JUDGE
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// freopen(\"input.txt\",\"r\",stdin);
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// #endif
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int T;cin>>T;while(T--)
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{
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Solve();
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}
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return 0;
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}
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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import random
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from bootcamp import Basebootcamp
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class Cnastiaandahiddenpermutationbootcamp(Basebootcamp):
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def __init__(self, min_n=3, max_n=10):
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self.min_n = min_n
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self.max_n = max_n
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def case_generator(self):
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n = random.randint(self.min_n, self.max_n)
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p = list(range(1, n + 1))
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random.shuffle(p)
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queries = self.simulate_queries(p)
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return {
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'n': n,
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'permutation': p,
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'queries': queries
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}
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def simulate_queries(self, p):
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queries = []
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n = len(p)
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def ask(t, i, j, x):
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i_index = i - 1 # Convert to 0-based
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j_index = j - 1
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pi = p[i_index]
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pj = p[j_index]
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if t == 1:
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res = max(min(x, pi), min(x + 1, pj))
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elif t == 2:
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res = min(max(x, pi), max(x + 1, pj))
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else:
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res = -1
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queries.append((t, i, j, x, res))
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return res
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a = [0] * n
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for i in range(0, n - 1, 2):
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x = ask(2, i + 1, (i + 1) + 1, 1)
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y = ask(1, i + 1, (i + 1) + 1, n - 1)
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if x == 2 and ask(1, i + 1, (i + 1) + 1, 1) == 1:
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a[i + 1] = 1
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if y == n - 1 and ask(2, i + 1, (i + 1) + 1, n - 1) == n:
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a[i] = n
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else:
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a[i] = y
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continue
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if y == n - 1 and ask(2, i + 1, (i + 1) + 1, n - 1) == n:
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a[i] = n
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a[i + 1] = x
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continue
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check = ask(2, (i + 1) + 1, i + 1, x)
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if check == x + 1:
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a[i] = x
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a[i + 1] = y
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else:
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a[i] = y
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a[i + 1] = x
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if n % 2 == 1:
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last = set(range(1, n + 1)) - set(a[:n-1])
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a[-1] = last.pop()
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return queries
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@staticmethod
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def prompt_func(question_case):
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queries = question_case['queries']
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queries_text = []
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for t, i, j, x, res in queries:
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queries_text.append(f"? {t} {i} {j} {x} → {res}")
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queries_str = "\n".join(queries_text)
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return f"""Nastia has a hidden permutation of length {question_case['n']}. The following queries were made and their responses are given:
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{queries_str}
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Determine the hidden permutation and provide your answer as "! p1 p2 ... pn" enclosed within [answer] tags. For example: [answer]! 1 2 3 4[/answer]"""
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@staticmethod
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def extract_output(output):
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import re
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matches = re.findall(r'!([\d\s]+)', output)
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if not matches:
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return None
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last_match = matches[-1].strip()
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try:
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solution = list(map(int, last_match.split()))
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except:
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return None
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return solution
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@classmethod
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def _verify_correction(cls, solution, identity):
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return solution == identity['permutation']
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