Permutation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
Special Judge

Problem Description

A permutation p1,p2,...,pn of 1,2,...,n is called a lucky permutation if and only if p≡0(mod|p−p|) for i=3...n.

Now you need to construct a lucky permutation with a given n.

Input

The first line is the number of test cases.

For each test case, one single line contains a positive integer n(3≤n≤105).

Output

For each test case, output a single line with n numbers p,p,...,p.

It is guaranteed that there exists at least one solution. And if there are different solutions, print any one of them.

Sample Input

Sample Output

1 3 2 6 4 5

Source

2017 ACM/ICPC 哈尔滨赛区网络赛——测试专用

题意就是给出的n，从1到n，满足条件p≡0(mod|p−p|) for i=3...n.

就是p%(p-p)==0就可以。

一开始想的好麻烦，队友太强啦，直接%1就可以啦，只要要计算的两个数相差为1就可以。

看代码就知道了，后面的数顺序和逆序都无所谓的。

代码:

 #include<bits/stdc++.h>

 using namespace std;

 const int N=1e5+;

 int a[N];

 int main(){

     int t,n;

     scanf("%d",&t);

     while(t--){

         scanf("%d",&n);

         int h=,k=n;

         for(int i=;i<=n;i+=)a[i]=h++;

         for(int i=;i<=n;i+=)a[i]=k--;

         for(int i=;i<=n;i++)

             printf("%d ",a[i]);

         printf("\n");

     }

     return ;

 }

队友太厉害啦，%%%。

permutation

HDU6235-Permutation-水题-2017中国大学生程序设计竞赛-哈尔滨站-重现赛

Permutation