CodeforcesRound#749(Div.1.2basedonTechnocup2022EliminationRound1)-D.Omkar and the Meaning of Life-题解

Posted 2021-11-08 Tisfy

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Codeforces Round #749 (Div. 1 + Div. 2, based on Technocup 2022 Elimination Round 1) - D. Omkar and the Meaning of Life

传送门
Time Limit: 2 seconds
Memory Limit: 256 megabytes

Problem Description

It turns out that the meaning of life is a permutation $p 1, p 2, \dots, p n$ of the integers $1, 2, \dots, n (2 \leq n \leq 100)$ . Omkar, having created all life, knows this permutation, and will allow you to figure it out using some queries.

A query consists of an array $a 1, a 2, \dots, a n$ of integers between $1$ and $n$ . a is not required to be a permutation. Omkar will first compute the pairwise sum of $a$ and $p$ , meaning that he will compute an array $s$ where $s_j=p_j+a_j$ for all $j = 1, 2, \dots, n$ . Then, he will find the smallest index $k$ such that sk occurs more than once in $s$ , and answer with $k$ . If there is no such index $k$ , then he will answer with $0$ .

You can perform at most $2 n$ queries. Figure out the meaning of life $p$ .

Interaction

Start the interaction by reading single integer $n (2 \leq n \leq 100)$ — the length of the permutation $p$ .

You can then make queries. A query consists of a single line “ $a_1\\ a_2\\ …\\ a_n$ ” $1≤a_j≤n)$ .

The answer to each query will be a single integer $k$ as described above $(0 \leq k \leq n)$ .

After making a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. 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.

To output your answer, print a single line “ $a_1\\ a_2\\ …\\ a_n$ ” then terminate.

You can make at most $2 n$ queries. Outputting the answer does not count as a query.

Hack Format

To hack, first output a line containing $n (2 \leq n \leq 100)$ , then output another line containing the hidden permutation $p_1,p_2,…,p_n$ of numbers from $1$ to $n$ .

Sample Input

Sample Onput


? 4 4 2 3 2

? 3 5 1 5 5

? 5 2 4 3 1

! 3 2 1 5 4

Note

In the sample, the hidden permutation $p$ is $[3, 2, 1, 5, 4]$ . Three queries were made.

The first query is $a = [4, 4, 2, 3, 2]$ . This yields $s = [3 + 4, 2 + 4, 1 + 2, 5 + 3, 4 + 2] = [7, 6, 3, 8, 6]$ . $6$ is the only number that appears more than once, and it appears first at index $2$ , making the answer to the query $2$ .

The second query is $a = [3, 5, 1, 5, 5]$ . This yields $s = [3 + 3, 2 + 5, 1 + 1, 5 + 5, 4 + 5] = [6, 7, 2, 10, 9]$ . There are no numbers that appear more than once here, so the answer to the query is 0.

The third query is $a = [5, 2, 4, 3, 1]$ . This yields $s = [3 + 5, 2 + 2, 1 + 4, 5 + 3, 4 + 1] = [8, 4, 5, 8, 5]$ . $5$ and $8$ both occur more than once here. $5$ first appears at index $3$ , while $8$ first appears at index $1$ , and $1 < 3$ , making the answer to the query $1$ .

Note that the sample is only meant to provide an example of how the interaction works; it is not guaranteed that the above queries represent a correct strategy with which to determine the answer.

题目大意

给你一个 $n$ ，这组数据其实是隐藏了 $n$ 的一个全排列。你可以进行最多 $2 n$ 次询问，每次输入一个新的序列，让隐藏的序列一一对应临时加上这个新的序列，之后把有相同的值的元素中，最小的下标返回给你。

让你通过询问，得出隐藏的序列。

比如样例隐藏了序列 $[3, 2, 1, 5, 4]$ ，你进行了一次询问并让隐藏序列一一对应地临时加上你给的序列 $[4, 4, 2, 3, 2]$ . 那么相加产生的临时序列是 $[3 + 4, 2 + 4, 1 + 2, 5 + 3, 4 + 2] = [7, 6, 3, 8, 6]$ . $6$ 不只一个，所有有相同值的元素集合是： $[第 2 个元素，第 5 个元素]$ ，其中下标最小的是第二个元素，它的下标是 $CodeforcesRound#749(Div.1.2basedonTechnocup2022EliminationRound1)-D.Omkar and the Meaning of Life-题解$

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