[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。

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

  2 

  3 #include <iostream>

  4 #include <cstdio>

  5 #include <algorithm>

  6 #include <cstdlib>

  7 #include <cstring>

  8 #include <map>

  9 #include <queue>

 10 #include <deque>

 11 #include <cmath>

 12 #include <vector>

 13 #include <ctime>

 14 #include <cctype>

 15 #include <set>

 16 #include <bitset>

 17 #include <functional>

 18 #include <numeric>

 19 #include <stdexcept>

 20 #include <utility>

 21 

 22 using namespace std;

 23 

 24 #define mem0(a) memset(a, 0, sizeof(a))

 25 #define mem_1(a) memset(a, -1, sizeof(a))

 26 #define lson l, m, rt << 1

 27 #define rson m + 1, r, rt << 1 | 1

 28 #define define_m int m = (l + r) >> 1

 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++)

 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)

 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)

 32 #define rep_down1(a, b) for (int a = b; a > 0; a--)

 33 #define all(a) (a).begin(), (a).end()

 34 #define lowbit(x) ((x) & (-(x)))

 35 #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) {}

 36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}

 37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}

 38 #define pchr(a) putchar(a)

 39 #define pstr(a) printf("%s", a)

 40 #define sstr(a) scanf("%s", a)

 41 #define sint(a) scanf("%d", &a)

 42 #define sint2(a, b) scanf("%d%d", &a, &b)

 43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)

 44 #define pint(a) printf("%d\n", a)

 45 #define test_print1(a) cout << "var1 = " << a << endl

 46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl

 47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl

 48 

 49 typedef long long LL;

 50 typedef pair<int, int> pii;

 51 typedef vector<int> vi;

 52 

 53 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};

 54 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };

 55 const int maxn = 2e2 + 7;

 56 const int md = 10007;

 57 const int inf = 1e9 + 7;

 58 const LL inf_L = 1e18 + 7;

 59 const double pi = acos(-1.0);

 60 const double eps = 1e-6;

 61 

 62 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}

 63 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}

 64 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}

 65 template<class T>T condition(bool f, T a, T b){return f?a:b;}

 66 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}

 67 int make_id(int x, int y, int n) { return x * n + y; }

 68 

 69 int p[maxn], f[maxn][maxn], ans[maxn];

 70 int main() {

 71     //freopen("in.txt", "r", stdin);

 72     int T, n, m;

 73     cin >> T;

 74     while (T--) {

 75         cin >> n >> m;

 76         rep_up0(i, m) {

 77             sint(p[i]);

 78         }

 79         mem0(f);

 80         f[n][0] = true;

 81         for (int i = n - 1; i; i --) {

 82             for (int j = 0; j < n - i + 1; j ++) {

 83                 rep_up0(k, m) {

 84                     if ((p[k] + n - i) % (n - i + 1) == j) continue;

 85                     int sta, tmp = p[k] % (n - i + 1);

 86                     if (j >= tmp) sta = j - tmp;

 87                     else sta = n - i + 1 - tmp + j;

 88                     f[i][j] = f[i][j] || f[i + 1][sta];

 89                 }

 90             }

 91         }

 92         int cnt = 0;

 93         rep_up0(i, n) {

 94             if (f[1][i]) ans[cnt ++] = i + 1;

 95         }

 96         cout << cnt << endl;

 97         rep_up0(i, cnt) {

 98             printf("%d%c", ans[i], i == cnt - 1? '\n' : ' ');

 99         }

100     }

101     return 0;

102 }

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