[hdu5218]DP-约瑟夫环变形

题意：n个人围成一圈，另外一个人最开始站在第一个人前面，每次从集合s里面随机选一个数x，这个人顺时针经过x个人后停下来，当前位置的前一个人出队，然后继续进行，求最后剩下的那个人的可能编号。

思路：由于只求最后一个人的编号，可以将一次操作后的人进行重编号，来进行状态转移，转化为子问题用dp来解决。dp方程比较容易写出，注意下细节就好了。

#pragma comment(linker, "/STACK:102400000,102400000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep_up0(a, b) for (int a = 0; a < (b); a++)
#define rep_up1(a, b) for (int a = 1; a <= (b); a++)
#define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
#define rep_down1(a, b) for (int a = b; a > 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sstr(a) scanf("%s", a)
#define sint(a) scanf("%d", &a)
#define sint2(a, b) scanf("%d%d", &a, &b)
#define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define pint(a) printf("%d
", a)
#define test_print1(a) cout << "var1 = " << a << endl
#define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
#define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl

typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
const int maxn = 2e2 + 7;
const int md = 10007;
const int inf = 1e9 + 7;
const LL inf_L = 1e18 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;

template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
template<class T>T condition(bool f, T a, T b){return f?a:b;}
template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
int make_id(int x, int y, int n) { return x * n + y; }

int p[maxn], f[maxn][maxn], ans[maxn];
int main() {
    //freopen("in.txt", "r", stdin);
    int T, n, m;
    cin >> T;
    while (T--) {
        cin >> n >> m;
        rep_up0(i, m) {
            sint(p[i]);
        }
        mem0(f);
        f[n][0] = true;
        for (int i = n - 1; i; i --) {
            for (int j = 0; j < n - i + 1; j ++) {
                rep_up0(k, m) {
                    if ((p[k] + n - i) % (n - i + 1) == j) continue;
                    int sta, tmp = p[k] % (n - i + 1);
                    if (j >= tmp) sta = j - tmp;
                    else sta = n - i + 1 - tmp + j;
                    f[i][j] = f[i][j] || f[i + 1][sta];
                }
            }
        }
        int cnt = 0;
        rep_up0(i, n) {
            if (f[1][i]) ans[cnt ++] = i + 1;
        }
        cout << cnt << endl;
        rep_up0(i, cnt) {
            printf("%d%c", ans[i], i == cnt - 1? '
' : ' ');
        }
    }
    return 0;
}