题目链接
题解
据说这题是多项式求逆
我太弱不会QAQ，只能\(O(nlog^2n)\)分治\(NTT\)设\(f[i]\)表示\(i\)个节点的简单无向连通图的数量
考虑转移，直接求不好求，我们知道\(n\)个点无向图的数量是\(2^{ {n \choose 2}}\)的，考虑用总数减去不连通的 既然图不连通，那么和\(1\)号点联通的点数一定小于\(n\)，我们枚举和\(1\)号点所在联通块大小，就可以得到式子：\[f[n] = 2^{ {n \choose 2}} - \sum\limits_{i = 1}^{n - 1}{n - 1 \choose i - 1}2^{ {n - i \choose 2}}f[i]\] 展开组合数变形得：\[f[n] = 2^{ {n \choose 2}} - (n - 1)!\sum\limits_{i = 1}^{n - 1}\frac{f[i] * i}{i!} * \frac{2^{n - i \choose 2}}{(n - i)!}\] 分治NTT即可#include#include #include #include #include #define REP(i,n) for (int i = 1; i <= (n); i++)#define cls(s) memset(s,0,sizeof(s))#define LL long long int#define res registerusing namespace std;const int maxn = 500000,maxm = 100005,INF = 1000000000,P = 1004535809;const int G = 3;int N,f[maxn],fac[maxn],fv[maxn],inv[maxn],C2[maxn];int A[maxn],B[maxn],R[maxn],w[2][maxn],L,n,m;inline int qpow(int a,LL b){ int ans = 1; for (; b; b >>= 1,a = 1ll * a * a % P) if (b & 1) ans = 1ll * ans * a % P; return ans;}inline void NTT(int* a,int f){ for (res int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]); for (res int i = 1; i < n; i <<= 1){ int gn = w[f][i]; for (res int j = 0; j < n; j += (i << 1)){ int g = 1,x,y; for (res int k = 0; k < i; k++,g = 1ll * g * gn % P){ x = a[j + k]; y = 1ll * g * a[j + k + i] % P; a[j + k] = (x + y) % P; a[j + k + i] = (x - y + P) % P; } } } if (f == 1) return; int nv = qpow(n,P - 2); for (res int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;}void solve(int l,int r){ if (l == r){ f[l] = ((C2[l] - 1ll * fac[l - 1] * f[l] % P) % P + P) % P; return; } int mid = l + r >> 1,t; solve(l,mid); m = (mid - l) + (r - l); L = 0; for (n = 1; n <= m; n <<= 1) L++; for (int i = 0; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1)); for (int i = 0; i < n; i++) A[i] = B[i] = 0; t = mid - l + 1; for (int i = 0; i < t; i++) A[i] = 1ll * f[l + i] * fv[l + i - 1] % P; t = r - l; B[0] = 0; for (int i = 1; i <= t; i++) B[i] = 1ll * C2[i] * fv[i] % P; NTT(A,1); NTT(B,1); for (int i = 0; i < n; i++) A[i] = 1ll * A[i] * B[i] % P; NTT(A,0); for (int i = mid + 1; i <= r; i++){ f[i] = (f[i] + A[i - l]) % P; } solve(mid + 1,r);}int main(){ scanf("%d",&N); fac[0] = fac[1] = inv[0] = inv[1] = fv[0] = fv[1] = 1; for (res int i = 2; i <= N; i++){ fac[i] = 1ll * fac[i - 1] * i % P; inv[i] = 1ll * (P - P / i) * inv[P % i] % P; fv[i] = 1ll * fv[i - 1] * inv[i] % P; } for (res int i = 1; i <= N; i++){ C2[i] = qpow(2,1ll * i * (i - 1) / 2); } for (res int i = 1; i < maxn; i <<= 1){ w[1][i] = qpow(G,(P - 1) / (i << 1)); w[0][i] = qpow(G,(-(P - 1) / (i << 1) % (P - 1) + (P - 1)) % (P - 1)); } solve(1,N); printf("%d\n",f[N]); return 0;}