HDU 6235.Permutation (2017中国大学生程序设计竞赛-哈尔滨站-重现赛)

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:

 #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，%%%。