[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\n", 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[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 2e2 + ;
 const int md = ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?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][] = true;
         for (int i = n - ; i; i --) {
             for (int j = ; j < n - i + ; j ++) {
                 rep_up0(k, m) {
                     if ((p[k] + n - i) % (n - i + ) == j) continue;
                     int sta, tmp = p[k] % (n - i + );
                     if (j >= tmp) sta = j - tmp;
                     else sta = n - i +  - tmp + j;
                     f[i][j] = f[i][j] || f[i + ][sta];
                 }
             }
         }
         int cnt = ;
         rep_up0(i, n) {
             if (f[][i]) ans[cnt ++] = i + ;
         }
         cout << cnt << endl;
         rep_up0(i, cnt) {
             printf("%d%c", ans[i], i == cnt - ? '\n' : ' ');
         }
     }
     return ;
 }