Cat, Fox and Double Maximum

题目描述

Fox loves permutations! She came up with the following problem and asked Cat to solve it:

You are given an even positive integer $n$ and a permutation${}^{†}$ $p$ of length $n$.

The score of another permutation $q$ of length $n$ is the number of local maximums in the array $a$ of length $n$, where $a_i=p_i+q_i$ for all $i$ ($1\le i\le n$). In other words, the score of $q$ is the number of $i$ such that $1<i<n$ (note the strict inequalities), $a_{i-1}<a_i$, and $a_i>a_{i+1}$ (once again, note the strict inequalities).

Find the permutation $q$ that achieves the maximum score for given $n$ and $p$. If there exist multiple such permutations, you can pick any of them.

${}^{†}$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).

输入描述

The first line of input contains an integer $t$ ($1\le t\le 10^4$) — the number of test cases in the input you will have to solve.

The first line of each test case contains one even integer $n$ ($4\le n\le 10^5$, $n$ is even) — the length of the permutation $p$.

The second line of each test case contains the $n$ integers $p_1,p_2,\dots ,p_n$ ($1\le p_i\le n$). It is guaranteed that $p$ is a permutation of length $n$.

It is guaranteed that the sum of $n$ across all test cases doesn’t exceed $10^5$.

输出描述

For each test case, output one line containing any permutation of length $n$ (the array $q$), such that $q$ maximizes the score under the given constraints.

样例输入 #1

4
4
1 2 3 4
4
4 3 1 2
6
6 5 1 4 2 3
8
1 2 4 5 7 6 8 3

样例输出 #1

2 4 1 3
3 1 4 2
2 5 1 4 3 6
5 4 8 2 7 1 6 3

原题

CF——传送门
洛谷——传送门

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;int main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t;cin >> t;while (t--){int n;cin >> n;vector<int> p(n);for (int i = 0; i < n; i++){cin >> p[i];}int n_pos = find(p.begin(), p.end(), n) - p.begin();vector<int> q(n);// 注意，由于我从下标0开始读入数据，所以在下面的描述中，0,2,4,6……是奇数位，而1,3,5,7……是偶数位if (n_pos & 1) // 如果值n在偶数位上，则将p排列翻转，使其变为奇数位上{reverse(p.begin(), p.end());}// n在偶数位和n在奇数位的情况最终都转化为n在奇数位的处理vector<pair<int, int>> even, odd; // 记录值和位置的数组// 先处理偶数位，偶数位需得到1~n/2，小的值交给大的for (int i = 1; i < n; i += 2){even.push_back({p[i], i});}sort(even.begin(), even.end(), greater<pair<int, int>>());// 按小值交给大值的顺序赋值1~n/2for (int i = 0; i < n / 2; i++){q[even[i].second] = i + 1;}// 再处理奇数位，奇数位的第一位得到n/2+1，剩余位得到n/2+2~n，大的值交给小的for (int i = 0; i < n; i += 2){odd.push_back({p[i], i});}sort(odd.begin() + 1, odd.end(), greater<pair<int, int>>());// 按大值交给小值的顺序赋值n/2+1~nfor (int i = 0; i < n / 2; i++){q[odd[i].second] = i + n / 2 + 1;}// 若之前翻转过p，则再翻转一次结果q就能得到所需答案if (n_pos & 1){reverse(q.begin(), q.end());}for (int i = 0; i < n; i++){cout << q[i] << " \n"[i == n - 1]; // 新学到的小寄巧()}}return 0;
}