E. Klever Permutation

链接：https://codeforces.com/problemset/problem/1927/E
思路：观察，可知每隔k个数据就是+1/-1，且间隔而分，思路如下：




然后按顺序打表就行
代码：

#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<sstream>
#include<string>
#include<string.h>
#include<iomanip>
#include<stdlib.h>
#include<map>
#include<queue>
#include<limits.h>
#include<climits>
#include<fstream>
#include<stack>
#include<set>

typedef long long ll;
using namespace std;
const int N = 2e5+10;
ll a[N];
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int t; cin >> t;
	while (t--)
	{
		int ptr = 1;
		int n, k; cin >> n >> k;
		for (int i = 1; i <= k; i += 2)//注意这个i+=2
		{
			for (int j = 0; i + j * k <= n; j++)
			{
				a[i + j * k] = ptr;
				ptr++;
			}
		}
		ptr = n;//从n开始倒着标
		for (int i = k; i > 0; i -= 2)//i-=2
		{
			for (int j = 0; i + j * k <= n; j++)
			{
				a[i + j * k] = ptr;
				ptr--;
			}
		}
		for (int i = 1; i <= n; i++)cout << a[i] << ' ';
		cout << endl;
	}

	return 0;
}