题目链接：https://codeforces.com/contest/1553/problem/E

题目大意：是否能在转置\(k\)轮和交换\(m\)次的条件下将\((1,2,3,...,n)\)变成\((a_{1},a_{2}，a_{3},...,a_{n})\)，\(0\leq k<n,0\leq m\leq \frac{n}{3}\)，转置指元素后移\(k\)位，交换指两个元素交换位置

题目思路：
先将每个数都减\(1\)
转置0轮\((0,1,2,3)\)
转置1轮\((3,0,1,2)\)
转置2轮\((2,3,0,1)\)
转置3轮\((1,2,3,0)\)
不难发现转置k轮实际上就是\(p[i] = i-k(mod \ n)\)
对于每一个转置我们可以求出对a序列来说不动点的个数
a序列为\((1,2,0,3)\)
转置0轮\((0,1,2,3)\)，不动点数\(cnt[0] = 1\)
转置1轮\((3,0,1,2)\)，不动点数\(cnt[1] = 0\)
转置2轮\((2,3,0,1)\)，不动点数\(cnt[2] = 1\)
转置3轮\((1,2,3,0)\)，不动点数\(cnt[3] = 2\)
不动点即\(a[i] =p[i] = i-k(mod \ n)\),得 \(k = i-a[i] (mod \ n)\)，因\(++cnt[k]\)即可
剩下都是需要移动的点，交换\(m\)次最多可导致\(2m\)个元素交换，因此可以判断如果\(cnt[k]+2m<n\)当前转置就不满足条件，跳过即可
因为\(m\leq \frac{n}{3}\)，最多可以交换\(\frac{2n}{3}\)个元素，剩下\(\frac{n}{3}\)是不动点的个数，又因为\(\sum cnt[k] = n\)，最多只有3个\(k\)是满足条件的，直接暴力即可
暴力求出环数\(num\)，需要移动的点即为\(n-num\)

AC代码：

#include <unordered_map>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <stack>
#include <deque>
#include <queue>
#include <cmath>
#include <map>
#include <set>
using namespace std;
typedef pair<int, int> PII;
typedef pair<double, int> PDI;
//typedef __int128 int128;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const int N = 3e5 + 10, M = 4e7 + 10;
const int base = 1e9;
const int P = 131;
const int mod = 1e9 + 7;
const double eps = 1e-9;
const double PI = acos(-1.0);
int a[N], cnt[N], ans[5];
int pos;
bool vis[N];
int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		pos = 0;
		int n, m;
		scanf("%d%d", &n, &m);
		for (int i = 0; i < n; ++i)
		{
			scanf("%d", &a[i]);
			--a[i];
			++cnt[(i - a[i] + n) % n];
		}
		for (int k = 0; k < n; ++k)
		{
			if (cnt[k] + 2 * m < n)
				continue;
			for (int i = 0; i < n; ++i)
				vis[i] = false;
			int num = 0;
			for (int i = 0; i < n; ++i)
			{
				if (vis[i])
					continue;
				for (int j = i; !vis[j]; j = (a[j] + k) % n)
					vis[j] = true;
				++num;
			}
			if (n - num <= m)
				ans[++pos] = k;
		}
		printf("%d ", pos);
		for (int i = 1; i <= pos; ++i)
			printf("%d ", ans[i]);
		printf("\n");
	}
	return 0;
}