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

Posted 2020-10-14 Persistent.

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_i≡0(mod|p_i?p_i?2|) 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 p₁,p₂,...,p_n.

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

 

 

Sample Input

1

6

 

 

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_i≡0(mod|p_i?p_i?2|) for i=3...n.

就是p_i%(p_i-p_i-2)==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:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=1e5+10;
 4 int a[N];
 5 int main(){
 6     int t,n;
 7     scanf("%d",&t);
 8     while(t--){
 9         scanf("%d",&n);
10         int h=1,k=n;
11         for(int i=1;i<=n;i+=2)a[i]=h++;
12         for(int i=2;i<=n;i+=2)a[i]=k--;
13         for(int i=1;i<=n;i++)
14             printf("%d ",a[i]);
15         printf("\n");
16     }
17     return 0;
18 }

 队友太厉害啦，%%%。