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+10;
int a[N];
int main(){
    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        int h=1,k=n;
        for(int i=1;i<=n;i+=2)a[i]=h++;
        for(int i=2;i<=n;i+=2)a[i]=k--;
        for(int i=1;i<=n;i++)
            printf("%d ",a[i]);
        printf("
");
    }
    return 0;
}

 队友太厉害啦，%%%。