# 【题目】K Derangement

AtCoder Regular Contest 144

C - K Derangement

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $600$ points

### Problem Statement

You are given positive integers $N$ and $K$. Find the lexicographically smallest permutation $A = (A_1, A_2, \ldots, A_N)$ of the integers from $1$ through $N$ that satisfies the following condition:

• $\lvert A_i - i\rvert \geq K$ for all $i$ ($1\leq i\leq N$).

If there is no such permutation, print -1.

### Constraints

• $2\leq N\leq 3\times 10^5$
• $1\leq K\leq N - 1$

### Input

Input is given from Standard Input in the following format:

$N$ $K$

### Output

Print the lexicographically smallest permutation $A = (A_1, A_2, \ldots, A_N)$ of the integers from $1$ through $N$ that satisfies the condition, in the following format:

$A_1$ $A_2$ $\ldots$ $A_N$

If there is no such permutation, print -1.

3 1

### Sample Output 1

2 3 1

Two permutations satisfy the condition: $(2, 3, 1)$ and $(3, 1, 2)$. For instance, the following holds for $(2, 3, 1)$:

• $\lvert A_1 - 1\rvert = 1 \geq K$
• $\lvert A_2 - 2\rvert = 1 \geq K$
• $\lvert A_3 - 3\rvert = 2 \geq K$

8 3

4 5 6 7 8 1 2 3

8 6

-1

### 解题笔记

• $N<2K$: 无解
• $2K\leq N\leq 3K$:

$$A(i) = \begin{cases} i + K & (1\leq i \leq N-K) \\ i - N + K & (N-K < i\leq N) \end{cases}$$

• $3K<N\leq 4K$:

$$A(i) = \begin{cases} i+K &(1\leq i \leq K)\\ i-K &(K<i\leq N-2K)\\ i + K &(N-2K< i\leq N-K)\\ i-2K &(N-K < i \leq 3K)\\ i-K &(3K < i \leq N) \end{cases}$$

• $4K<N$:

$$A(i) = \begin{cases} i + K & (1\leq i \leq K) \\ i - K & (K < i\leq 2K) \\ A(i) & (2K<i\leq N) \end{cases}$$

这个意思是前 $2K$ 位为 $(K+1, \ldots, 2K, 1, \ldots, K)$，剩余的部分的 $N'=N-2K$，再选择符合 $N'$ 大小的公式。不断使用情况四的公式进行递推，一定可以转化为情况二或情况三。

### 代码

#include <bits/stdc++.h>

using namespace std;

int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int N, K;
cin >> N >> K;
if (2 * K > N)
{
cout << -1 << endl;
return 0;
}
int cnt = 0;
while (N - cnt >= 4 * K)
{
for (int i = 1; i <= K; i++)
cout << cnt + i + K << ' ';
for (int i = 1; i <= K; i++)
cout << cnt + i << ' ';
cnt += 2 * K;
}
if (N - cnt <= 3 * K)
{
int i = 1;
for (; i <= (N - cnt) - K; i++)
cout << i + K + cnt << ' ';
for (; i <= (N - cnt); i++)
cout << i - (N - cnt) + K + cnt << ' ';
}
else
{
int i = 1;
for (; i <= K; i++)
cout << i + K + cnt << ' ';
for (; i <= (N - cnt) - 2 * K; i++)
cout << i - K + cnt << ' ';
for (; i <= (N - cnt) - K; i++)
cout << i + K + cnt << ' ';
for (; i <= 3 * K; i++)
cout << i - 2 * K + cnt << ' ';
for (; i <= (N - cnt); i++)
cout << i - K + cnt << ' ';
}
cout << endl;
return 0;
}