BZOJ5093php
點之間是沒有區別的,因此咱們能夠計算出一個點的全部貢獻,而後乘上\(n\)
一個點可能向剩餘的\(n - 1\)個點連邊,那麼就有
\[ans = 2^{{n - 1 \choose 2}}n \sum\limits_{i = 0}^{n - 1} {n - 1 \choose i} i^k\]
顯然要求
\[\sum\limits_{i = 0}^{n} {n \choose i} i^k\]
而後我就不知道怎麼作了。。ios
翻翻題解
有這樣一個結論:
\[n^k = \sum\limits_{i = 0}^{k} \begin{Bmatrix} k \\ i \end{Bmatrix} {n \choose i} i!\]
那麼就有
\[ \begin{aligned} \sum\limits_{i = 0}^{n} {n \choose i} i^k &= \sum\limits_{i = 0}^{n} {n \choose i} \sum\limits_{j = 0}^{i} \begin{Bmatrix} k \\ j \end{Bmatrix} {i \choose j}j! \\ &= \sum\limits_{j = 0}^{n}\begin{Bmatrix} k \\ j \end{Bmatrix} j! \sum\limits_{i = j}^{n} {n \choose i} {i \choose j} \\ &= \sum\limits_{j = 0}^{n}\begin{Bmatrix} k \\ j \end{Bmatrix} j! {n \choose j} 2^{n - j} \\ \end{aligned} \]
解釋一下最後一步
\[\sum\limits_{i = j}^{n} {n \choose i} {i \choose j}\]
直觀來看是從\(n\)中取出\(i\)個,而後從\(i\)中取出\(j\)個
實際上等價於從\(n\)中取出\(j\)個,剩餘隨便取c++
最後只須要求出第二類斯特林數,用第二類斯特林反演便可
\[ \begin{aligned} \begin{Bmatrix} n \\ m \end{Bmatrix} &= \frac{1}{m!} \sum\limits_{i = 0}^{m} (-1)^{i}{m \choose i}(m - i)^{n} \\ &= \sum\limits_{i = 0}^{m} \frac{(-1)^{i}}{i!} \times \frac{(m - i)^{n}}{(m - i)!} \\ \end{aligned} \]
\(NTT\)便可spa
#include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<map> #define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt) #define REP(i,n) for (int i = 1; i <= (n); i++) #define mp(a,b) make_pair<int,int>(a,b) #define cls(s) memset(s,0,sizeof(s)) #define cp pair<int,int> #define LL long long int using namespace std; const int maxn = 800005,maxm = 100005,INF = 1000000000; inline int read(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();} return out * flag; } const int G = 3,P = 998244353; inline int qpow(int a,LL b){ int re = 1; for (; b; b >>= 1,a = 1ll * a * a % P) if (b & 1) re = 1ll * re * a % P; return re; } int R[maxn]; void NTT(int* a,int n,int f){ for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]); for (int i = 1; i < n; i <<= 1){ int gn = qpow(G,(P - 1) / (i << 1)); for (int j = 0; j < n; j += (i << 1)){ int g = 1,x,y; for (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) % P; } } } if (f == 1) return; int nv = qpow(n,P - 2); reverse(a + 1,a + n); for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P; } int fac[maxn],inv[maxn],fv[maxn],C[maxn]; int S[maxn],B[maxn],N,K; void init(){ fac[0] = fac[1] = inv[0] = inv[1] = fv[0] = fv[1] = 1; for (int i = 2; i <= K; 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; } C[0] = 1; int E = min(N - 1,K); for (int i = 1; i <= E; i++) C[i] = 1ll * C[i - 1] * (N - i) % P * inv[i] % P; } void work(){ int n = 1,L = 0; for (int i = 0; i <= K; i++){ S[i] = (((i & 1) ? -1 : 1) * fv[i] + P) % P; B[i] = 1ll * qpow(i,K) * fv[i] % P; } while (n <= (K << 1)) n <<= 1,L++; for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1)); NTT(S,n,1); NTT(B,n,1); for (int i = 0; i < n; i++) S[i] = 1ll * S[i] * B[i] % P; NTT(S,n,-1); //REP(i,10) printf("%d ",S[i]); puts(""); int ans = 0; for (int i = 0; i < N; i++){ if (i > K) break; ans = (ans + 1ll * S[i] * fac[i] % P * C[i] % P * qpow(2,N - 1 - i) % P) % P; } ans = 1ll * ans * N % P * qpow(2,1ll * (N - 1) * (N - 2) / 2) % P; printf("%d\n",ans); } int main(){ N = read(); K = read(); if (N == 1){puts("0"); return 0;} if (N == 2){puts("2"); return 0;} init(); work(); return 0; }