Codeforces #663 (Div. 2) A. Suborrays

A. Suborrays

原题链接：https://codeforces.com/contest/1391/problem/A

time limit per test：1 second memory limit per test：256 megabytes input：standard input output：standard output

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 $)$.

For a positive integer $n$, we call a permutation p of length n good if the following condition holds for every pair $i$ and $j$ $(1≤i≤j≤n)$—

• $(p_i$ $OR$ $p_{i+1}$ $OR$ $…$ $OR$ $p_{j−1}$ $OR$ $p_j)≥j−i+1$ , where $OR$ denotes the bitwise $OR$ operation.

In other words, a permutation $p$ is good if for every subarray of $p$, the $OR$ of all elements in it is not less than the number of elements in that subarray.

Given a positive integer $n$, output any good permutation of length $n$. We can show that for the given constraints such a permutation always exists.

Input
Each test contains multiple test cases. The first line contains the number of test cases $t (1≤t≤100)$. Description of the test cases follows.

The first and only line of every test case contains a single integer $n (1≤n≤100)$.

Output
For every test, output any good permutation of length n on a separate line.

Example
input

3
1
3
7

output

1
3 1 2
4 3 5 2 7 1 6

Note
For n = 3, $[3,1,2]$ is a good permutation. Some of the subarrays are listed below.

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• $3$ $OR$ $1$ $=3≥2 (i=1,j=2)$
• $3$ $OR$ $1$ $OR$ $2=3≥3 (i=1,j=3)$
• $1$ $OR$ $2=3≥2 (i=2,j=3)$
• $1≥1 (i=2,j=2)$
  Similarly, you can verify that $[4,3,5,2,7,1,6]$ is also good.

题意：输入一个长度n，用 $1$ ~ $n$的数字进行随机排列。定义对于一个正整数n，如果对于每一对 $i$ 和 $j(1≤i≤j≤n)$都满足，对于p的每个子数组，其中所有元素的OR不小于该子数列中元素的个数。我们称长度为 $n$ 的排列 $p$ 是好的。
题解：两个数进行位运算了，且范围是1~n，那么进行位运算后的结果不会小于元素的个数。简而言之就是：无论怎么输出这个置换序列都是正确的。（我是逆序输出的。）

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		int n;
		cin >> n;
		for (int i = n; i >= 1; i--)
			printf("%d ",i);
		cout << endl;
	}
	return 0;
}

其实这次的div2做的有点像div3的难度。