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 $p1,p2,\dots ,pn$ of the integers $1,2,\dots ,n(2\le n\le 100)$. 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 $a1,a2,\dots ,an$ 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 $2n$ queries. Figure out the meaning of life $p$.

Interaction

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

You can then make queries. A query consists of a single line “$?a_1{a}_2\dots a_n$” $(1\le a_j\le n)$.

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

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 “$!a_1{a}_2\dots a_n$” then terminate.

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

Hack Format

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

Sample Input

5201

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.

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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$2n$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$2\ast n$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。

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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;
}

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Tisfy：https://letmefly.blog.csdn.net/article/details/120837226

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