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。
1 #pragma comment(linker, "/STACK:10240000,10240000") 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 <stack> 16 #include <set> 17 #include <bitset> 18 #include <functional> 19 #include <numeric> 20 #include <stdexcept> 21 #include <utility> 22 23 using namespace std; 24 25 #define mem0(a) memset(a, 0, sizeof(a)) 26 #define mem_1(a) memset(a, -1, sizeof(a)) 27 #define lson l, m, rt << 1 28 #define rson m + 1, r, rt << 1 | 1 29 #define define_m int m = (l + r) >> 1 30 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 31 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 32 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 33 #define rep_down1(a, b) for (int a = b; a > 0; a--) 34 #define all(a) (a).begin(), (a).end() 35 #define lowbit(x) ((x) & (-(x))) 36 #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) {} 37 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 38 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 39 #define pchr(a) putchar(a) 40 #define pstr(a) printf("%s", a) 41 #define sstr(a) scanf("%s", a) 42 #define sint(a) scanf("%d", &a) 43 #define sint2(a, b) scanf("%d%d", &a, &b) 44 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) 45 #define pint(a) printf("%d ", a) 46 #define test_print1(a) cout << "var1 = " << a << endl 47 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl 48 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl 49 #define mp(a, b) make_pair(a, b) 50 #define pb(a) push_back(a) 51 52 typedef long long LL; 53 typedef pair<int, int> pii; 54 typedef vector<int> vi; 55 56 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; 57 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; 58 const int maxn = 1e4 + 7; 59 const int md = 1e9 + 7; 60 const int inf = 1e9 + 7; 61 const LL inf_L = 1e18 + 7; 62 const double pi = acos(-1.0); 63 const double eps = 1e-6; 64 65 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 66 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;} 67 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;} 68 template<class T>T condition(bool f, T a, T b){return f?a:b;} 69 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 70 int make_id(int x, int y, int n) { return x * n + y; } 71 72 struct Graph { 73 vector<vector<int> > G; 74 void clear() { G.clear(); } 75 void resize(int n) { G.resize(n + 2); } 76 void add(int u, int v) { G[u].push_back(v); } 77 vector<int> & operator [] (int u) { return G[u]; } 78 }; 79 Graph G; 80 int n, m; 81 int pre[maxn], lowlink[maxn], sccno[maxn], dfs_clock, scc_cnt; 82 stack<int> S; 83 int cnt[maxn], a[maxn]; 84 85 void dfs(int u) { 86 pre[u] = lowlink[u] = ++ dfs_clock; 87 S.push(u); 88 rep_up0(i, G[u].size()) { 89 int v = G[u][i]; 90 if (!pre[v]) { 91 dfs(v); 92 min_update(lowlink[u], lowlink[v]); 93 } 94 else if (!sccno[v]) { 95 min_update(lowlink[u], pre[v]); 96 } 97 } 98 if (lowlink[u] == pre[u]) { 99 scc_cnt ++; 100 for(;; ) { 101 int x = S.top(); S.pop(); 102 sccno[x] = scc_cnt; 103 if (x == u) break; 104 } 105 } 106 } 107 int find_scc(int n) { 108 dfs_clock = scc_cnt = 0; 109 mem0(sccno); 110 mem0(pre); 111 rep_up0(i, n) { 112 if (!pre[i]) dfs(i); 113 } 114 mem0(cnt); 115 int c = 0; 116 rep_up0(i, n) { 117 cnt[sccno[i]] ++; 118 } 119 rep_up1(i, scc_cnt) { 120 if (cnt[i] >= 2) c ++; 121 } 122 rep_up0(i, n) if (G[i].size() == 0) c ++; 123 int ans = 1; 124 rep_up0(i, c) { 125 ans = (ans << 1) % md; 126 } 127 return ans; 128 } 129 130 int P(int x) { 131 int ans = 0; 132 rep_up0(i, m + 1) { 133 ans = (ans * x + a[m - i]) % n; 134 } 135 return ans; 136 } 137 138 int main() { 139 //freopen("in.txt", "r", stdin); 140 int T; 141 cin >> T; 142 while (T --) { 143 cin >> n >> m; 144 G.clear(); 145 G.resize(n); 146 rep_up0(i, m + 1) sint(a[i]); 147 rep_up0(i, n) { 148 int x = P(i); 149 if (x != i) G.add(i, x); 150 } 151 cout << find_scc(n) << endl; 152 } 153 return 0; 154 }