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

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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 $2$ 。因此，你得到了询问的结果： $2$ 。

解题思路

其实每次询问可以把其中一个数 $a$ 加一，其他所有数都加2。那么如果有 $b = a - 1$ ，那么因为 $a$ 加了 $1$ 而 $b$ 加了 $2$ ，所以 $a$ 和 $b$ 加上临时序列后的值就会相同，因此就会返回这两个数中下标最小的元素。如果很荣幸 $b = a - 1$ 在 $b$ 的前面，那么我们就得到了比 $a$ 小 $1$ 的那个数的下标。

同理，只把 $a$ 加上 $2$ ，其他元素都加上 $1$ ，就可以令 $a$ 和 $a + 1$ 相等（如果存在的话）。如果 $a + 1$ 在 $a$ 的前面，那么我们就能得到比 $a$ 大 $1$ 的数的下标。

正好 $2 * n$ 次询问，我们可以得到 $n - 1$ 个“大1关系”，即得到了 $n - 1$ 对相差为 $1$ 的数。

因为没有 $0$ ，所以不存在比 $1$ 小 $1$ 的数，由此特殊条件我们可以判断出最小的数 $1$ 所在的位置。之后依据 $n - 1$ 对“大1关系”，就可以还原出原始数组的样子。

AC代码

/** @Author: LetMeFly* @Date: 2021-10-17 20:10:26* @LastEditors: LetMeFly* @LastEditTime: 2021-10-19 00:07:29*/ // LastEditTime: 2021-10-17 20:59:28
#include <bits/stdc++.h>
using namespace std;
#define mem(a) memset(a, 0, sizeof(a))
#define dbg(x) cout << #x << " = " << x << endl
#define fi(i, l, r) for (int i = l; i < r; i++)
#define cd(a) scanf("%d", &a)
typedef long long ll;
struct Conditions {int l, r; // 下标为l的元素比下标为r的大1// bool xd;
} cond[1010];
int sx[1010] = {0}; // sx[i]用来记录下标为i的数是否当过“比其他数大1的数”。如果没有，那么下标为i的数就是1
int Next[1010]; // Next[i]代表下标为i的数+1的数的下标
int ans[1010]; // 答案。ans[i]代表下标为i的数的真实隐藏的数
int main() {int n;cin >> n; // n个数int cnt = 0; // 现在又了cnt个“大1关系”for (int i = 1; i <= n; i++) {printf("? "); // 询问for (int j = 1; j <= n; j++) {if (j != i) printf("2 "); // 这个数之外的数都加2else printf("1 "); // 这个数+1}puts(""); // 换行fflush(stdout); // 清空缓冲区int answer; // 这个询问的回答cd(answer); // scanf("%d", &answer);if (answer != 0 && answer < i) { // answer=0代表加上后没有相同的数，answer=i说明要找的数在i后面cond[cnt].l = answer; // 回答的数的下标cond[cnt].r = i; // 比只加了1的数 小1// cond[cnt].xd = true;cnt++; // 已有的“大1关系”的数++}printf("? "); // 询问for (int j = 1; j <= n; j++) {if (j != i) printf("1 "); // 其他数+2else printf("2 "); // 这个数+1}puts(""); // 换行fflush(stdout); // 清空缓冲区cd(answer); // 同上if (answer != 0 && answer < i) {cond[cnt].l = i;cond[cnt].r = answer;// cond[cnt].xd = false;cnt++;}}for (int i = 0; i < cnt; i++) {Next[cond[i].l] = cond[i].r; // 下标为cond[i].l的数 +1得到的数 的下标是cond[i].rsx[cond[i].r]++; // cond[i].r当过“比其他数大1的数”}int one; // 1所在的下标for (int i = 1; i <= n; i++) {if (!sx[i]) { // 这个数没当过“比其他数大1的数”one = i; // 1就是它了break;}}for (int i = 1; i <= n; i++) { // 一共要还原n个数ans[one] = i; // 答案（原始的隐藏的数）one = Next[one]; // 比它大1的数}printf("! "); // 输出最终答案for (int i = 1; i <= n; i++) {printf("%d ", ans[i]);}puts(""); // 换行return 0;
}