csu1617]强连通分量

题意：定义域属于一个集合S={0,1,...,n-1}，求S的子集个数，满足以子集的元素为定义域的函数P(x)的值域等于子集本身。

思路：以元素为点，x到P(x)连一条有向边，不难发现，如果有一个有向环，那么环上的元素构成的集合就满足要求。所以问题转化为求有向环的个数，由于有向环之间不可能有交点（同一个点有且仅有一条出边），所以答案就是2^有向环的个数（如果选了有向环上的一点，那么整个有向环必须全部选）。所以只要用tarjan算法统计点数大于等于2的强连通分量个数然后加上自环的，就得到了有向环的个数了。

 #pragma comment(linker, "/STACK:10240000,10240000")

 #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 <stack>
 #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
 #define mp(a, b) make_pair(a, b)
 #define pb(a) push_back(a)

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

 const int dx[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 1e4 + ;
 const int md = 1e9 + ;
 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; }

 struct Graph {
     vector<vector<int> > G;
     void clear() { G.clear(); }
     void resize(int n) { G.resize(n + ); }
     void add(int u, int v) { G[u].push_back(v); }
     vector<int> & operator [] (int u) { return G[u]; }
 };
 Graph G;
 int n, m;
 int pre[maxn], lowlink[maxn], sccno[maxn], dfs_clock, scc_cnt;
 stack<int> S;
 int cnt[maxn], a[maxn];

 void dfs(int u) {
     pre[u] = lowlink[u] = ++ dfs_clock;
     S.push(u);
     rep_up0(i, G[u].size()) {
         int v = G[u][i];
         if (!pre[v]) {
             dfs(v);
             min_update(lowlink[u], lowlink[v]);
         }
         else if (!sccno[v]) {
             min_update(lowlink[u], pre[v]);
         }
     }
     if (lowlink[u] == pre[u]) {
         scc_cnt ++;
         for(;; ) {
             int x = S.top(); S.pop();
             sccno[x] = scc_cnt;
             if (x == u) break;
         }
     }
 }
 int find_scc(int n) {
     dfs_clock = scc_cnt = ;
     mem0(sccno);
     mem0(pre);
     rep_up0(i, n) {
         if (!pre[i]) dfs(i);
     }
     mem0(cnt);
     int c = ;
     rep_up0(i, n) {
         cnt[sccno[i]] ++;
     }
     rep_up1(i, scc_cnt) {
         if (cnt[i] >= ) c ++;
     }
     rep_up0(i, n) if (G[i].size() == ) c ++;
     int ans = ;
     rep_up0(i, c) {
         ans = (ans << ) % md;
     }
     return ans;
 }

 int P(int x) {
     int ans = ;
     rep_up0(i, m + ) {
         ans = (ans * x + a[m - i]) % n;
     }
     return ans;
 }

 int main() {
     //freopen("in.txt", "r", stdin);
     int T;
     cin >> T;
     while (T --) {
         cin >> n >> m;
         G.clear();
         G.resize(n);
         rep_up0(i, m + ) sint(a[i]);
         rep_up0(i, n) {
             int x = P(i);
             if (x != i) G.add(i, x);
         }
         cout << find_scc(n) << endl;
     }
     return ;
 }