#D7561. Median Queries
Median Queries
Median Queries
This is an interactive problem.
There is a secret permutation p (1-indexed) of numbers from 1 to n. More formally, for 1 ≤ i ≤ n, 1 ≤ p[i] ≤ n and for 1 ≤ i < j ≤ n, p[i] ≠ p[j]. It is known that p[1]<p[2].
In 1 query, you give 3 distinct integers a,b,c (1 ≤ a,b,c ≤ n), and receive the median of {|p[a]-p[b]|,|p[b]-p[c]|,|p[a]-p[c]|}.
In this case, the median is the 2-nd element (1-indexed) of the sequence when sorted in non-decreasing order. The median of {4,6,2} is 4 and the median of {0,123,33} is 33.
Can you find the secret permutation in not more than 2n+420 queries?
Note: the grader is not adaptive: the permutation is fixed before any queries are made.
Input
The first line of input contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
The first line of each testcase consists of a single integer n (20 ≤ n ≤ 100000) — the length of the secret permutation.
It is guaranteed that the sum of n over all test cases does not exceed 100000.
Interaction
For each testcase, you begin the interaction by reading n.
To perform a query, output "? a b c" where a,b,c is the 3 indices you want to use for the query.
Numbers have to satisfy 1 ≤ a,b,c ≤ n and a ≠ b,b ≠ c,a ≠ c.
For each query, you will receive a single integer x: the median of {|p[a]-p[b]|,|p[b]-p[c]|,|p[a]-p[c]|}.
In case your query is invalid or you asked more than 2n+420 queries, the interactor will print "−1" and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.
When you have determined the secret permutation, output "! p[1] p[2] ... p[n]". If the secret permutation is correct, the interactor will print "1". Otherwise, the interactor will print "-1" and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.
After printing a query do not forget to output the end of line and flush the output. Otherwise, you will get Idleness limit exceeded verdict.
To do this, use:
- fflush(stdout) or cout.flush() in C++;
- System.out.flush() in Java;
- flush(output) in Pascal;
- stdout.flush() in Python;
- see documentation for other languages.
Hacks:
To hack, use the following format of test:
The first line should contain a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
The first line of each testcase should contain a single integer n (20 ≤ n ≤ 100000) — the length of the secret permutation.
The following line of should contain n integers p[1],p[2],p[3],…,p[n]. p[1]<p[2] and p must be a permutation of integers from 1 to n.
You must ensure that the sum of n over all testcases does not exceed 100000.
Example
Input
1 20
6
9
1
Output
? 1 5 2
? 20 19 2
! 9 10 19 7 16 18 11 14 15 6 20 8 17 4 5 3 12 2 13 1
Note
The secret permutation is {9,10,19,7,16,18,11,14,15,6,20,8,17,4,5,3,12,2,13,1}.
For the first query, the values of (a,b,c) is (1,5,2). Since p[1]=9, p[5]=16 and p[2]=10. The return value is the median of {|9-16|,|16-10|,|9-10|} which is 6.
For the second query, the values of (a,b,c) is (20,19,2). Since p[20]=1, p[19]=13 and p[2]=10. The return value is the median of {|1-13|,|13-10|,|1-10|} which is 9.
By some miracle, we have figured out that the secret permutation is {9,10,19,7,16,18,11,14,15,6,20,8,17,4,5,3,12,2,13,1}. We output it and receive 1 from the interactor, meaning that we have guessed the secret permutation correctly.
inputFormat
Input
The first line of input contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
The first line of each testcase consists of a single integer n (20 ≤ n ≤ 100000) — the length of the secret permutation.
It is guaranteed that the sum of n over all test cases does not exceed 100000.
Interaction
For each testcase, you begin the interaction by reading n.
To perform a query,
outputFormat
output "? a b c" where a,b,c is the 3 indices you want to use for the query.
Numbers have to satisfy 1 ≤ a,b,c ≤ n and a ≠ b,b ≠ c,a ≠ c.
For each query, you will receive a single integer x: the median of {|p[a]-p[b]|,|p[b]-p[c]|,|p[a]-p[c]|}.
In case your query is invalid or you asked more than 2n+420 queries, the interactor will print "−1" and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.
When you have determined the secret permutation, output "! p[1] p[2] ... p[n]". If the secret permutation is correct, the interactor will print "1". Otherwise, the interactor will print "-1" and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.
After printing a query do not forget to output the end of line and flush the output. Otherwise, you will get Idleness limit exceeded verdict.
To do this, use:
- fflush(stdout) or cout.flush() in C++;
- System.out.flush() in Java;
- flush(output) in Pascal;
- stdout.flush() in Python;
- see documentation for other languages.
Hacks:
To hack, use the following format of test:
The first line should contain a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
The first line of each testcase should contain a single integer n (20 ≤ n ≤ 100000) — the length of the secret permutation.
The following line of should contain n integers p[1],p[2],p[3],…,p[n]. p[1]<p[2] and p must be a permutation of integers from 1 to n.
You must ensure that the sum of n over all testcases does not exceed 100000.
Example
Input
1 20
6
9
1
Output
? 1 5 2
? 20 19 2
! 9 10 19 7 16 18 11 14 15 6 20 8 17 4 5 3 12 2 13 1
Note
The secret permutation is {9,10,19,7,16,18,11,14,15,6,20,8,17,4,5,3,12,2,13,1}.
For the first query, the values of (a,b,c) is (1,5,2). Since p[1]=9, p[5]=16 and p[2]=10. The return value is the median of {|9-16|,|16-10|,|9-10|} which is 6.
For the second query, the values of (a,b,c) is (20,19,2). Since p[20]=1, p[19]=13 and p[2]=10. The return value is the median of {|1-13|,|13-10|,|1-10|} which is 9.
By some miracle, we have figured out that the secret permutation is {9,10,19,7,16,18,11,14,15,6,20,8,17,4,5,3,12,2,13,1}. We output it and receive 1 from the interactor, meaning that we have guessed the secret permutation correctly.
样例
1
20
6
9
1
? 1 5 2
? 20 19 2
! 9 10 19 7 16 18 11 14 15 6 20 8 17 4 5 3 12 2 13 1