题意:
给定\(n\)个点,一个图的价值定义为所有点的度数的\(k\)次方之和。
现在计算所有\(n\)个点的简单无向图的价值之和。
思路:
将式子列出来:
\[
\sum_{i=1}^n\sum_{j=0}^{n-1}{n-1\choose j}2^{\frac{(n-1)(n-2)}{2}}j^k
\]
表示分别考虑每个点的贡献,我们只需要枚举其度数即可,其余的边任意连。
然后我们将后面的\(j^k\)用第二类斯特林数展开:
\[
\begin{aligned}
&\sum_{i=1}^{n}\sum_{j=0}^{n-1}{n-1\choose j}2^{\frac{(n-1)(n-2)}{2}}\sum_{t=1}^{j}{j\choose t}t!\begin{Bmatrix}
k \\ t
\end{Bmatrix}\ =&\sum_{i=1}^n2^{\frac{(n-1)(n-2)}{2}}\sum_{t=1}^{n-1}t!\begin{Bmatrix}
k \\ t
\end{Bmatrix}\sum_{j=t}^{n-1}{n - 1 \choose j}{j \choose t}\ =&\sum_{i=1}^n2^{\frac{(n-1)(n-2)}{2}}\sum_{t=1}^{n-1}t!\begin{Bmatrix}
k \\ t
\end{Bmatrix}\sum_{j=t}^{n-1}{n - 1 \choose j}{j \choose t}\ =&\sum_{i=1}^n2^{\frac{(n-1)(n-2)}{2}}\sum_{t=1}^{n-1}t!{n-1\choose t}\begin{Bmatrix}
k \\ t
\end{Bmatrix}\sum_{j=t}^{n-1}{n - 1 - t\choose j - t}\ =&\sum_{i=1}^n2^{\frac{(n-1)(n-2)}{2}}\cdot\sum_{t=1}^{n-1}2^{n-1-t}t!{n-1\choose t}\begin{Bmatrix}
k \\ t
\end{Bmatrix}\ =&\sum_{i=1}^n2^{\frac{(n-1)(n-2)}{2}}(n-1)!\cdot\sum_{t=1}^{n-1}\frac{2^{n-1-t}}{(n-1-t)!}\begin{Bmatrix}
k \\ t
\end{Bmatrix}
\end{aligned}
\]
似乎可以不要最后一行,对于每个点预处理之后可以直接\(O(n)\)算了。
因为卷积系数要求\(\displaystyle \begin{Bmatrix} k \\ t \end{Bmatrix}\),注意到这是一行的第二类斯特林数,那么我们可以直接通过\(FFT\)在\(O(klogk)\)的时间内预处理出来。预处理详见:传送门。
代码如下:
/*
* Author: heyuhhh
* Created Time: 2019/12/11 22:57:14
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#define MP make_pair
#define fi first
#define se second
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << '\n'; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
#else
#define dbg(...)
#endif
void pt() {std::cout << '\n'; }
template<typename T, typename...Args>
void pt(T a, Args...args) {std::cout << a << ' '; pt(args...); }
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 2e5 + 5, MOD = 998244353;
int n, k, m;
ll qpow(ll a, ll b) {
ll ans = 1;
while(b) {
if(b & 1) ans = ans * a % MOD;
a = a * a % MOD;
b >>= 1;
}
return ans;
}
const int P = 998244353, G = 3, Gi = 332748118;
int lim = 1, L, r[N * 4];
ll a[N * 4], b[N * 4];
void NTT(ll *A, int type) {
for(int i = 0; i < lim; i++)
if(i < r[i]) swap(A[i], A[r[i]]);
for(int mid = 1; mid < lim; mid <<= 1) {
ll Wn = qpow( type == 1 ? G : Gi , (P - 1) / (mid << 1)); //Wn = g ^ ((p - 1) / n) (mod p)
for(int j = 0; j < lim; j += (mid << 1)) {
ll w = 1;
for(int k = 0; k < mid; k++, w = (w * Wn) % P) {
int x = A[j + k], y = w * A[j + k + mid] % P;
A[j + k] = (x + y) % P,
A[j + k + mid] = (x - y + P) % P;
}
}
}
}
void solve(ll *a, ll *b) {
while(lim <= m + m) lim <<= 1, L++;
for(int i = 0; i < lim; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (L - 1));
for(int i = n + 1; i < lim; i++) a[i] = 0; //a,b need init
for(int i = m + 1; i < lim; i++) b[i] = 0;
NTT(a, 1); NTT(b, 1);
for(int i = 0; i < lim; i++) a[i] = (a[i] * b[i]) % P;
NTT(a, -1);
ll inv = qpow(lim, P - 2);
for(int i = 0; i < lim; i++) a[i] = a[i] * inv % P;
}
int fac[N], inv[N], c[N];
void init() {
fac[0] = 1;
for(int i = 1; i < N; i++) fac[i] = 1ll * fac[i - 1] * i % MOD;
inv[N - 1] = qpow(fac[N - 1], MOD - 2);
for(int i = N - 2; i >= 0; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % MOD;
c[0] = 1;
for(int i = 1; i < N; i++) c[i] = 1ll * c[i - 1] * (n - i) % MOD * qpow(i, MOD - 2) % MOD;
m = k;
for(int i = 0; i <= m; i++) {
a[i] = (i & 1) ? MOD - inv[i] : inv[i];
b[i] = qpow(i, k) * inv[i] % MOD;
}
solve(a, b);
}
void run(){
ll ans = 1ll * n * qpow(2, 1ll * (n - 1) * (n - 2) / 2) % MOD;
ll res = 0;
for(int i = 0; i <= m; i++) {
res = (res + a[i] * fac[i] % MOD * c[i] % MOD * qpow(2, n - i - 1) % MOD) % MOD;
}
ans = ans * res % MOD;
cout << ans << '\n';
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
cin >> n >> k;
init();
run();
return 0;
}
原文地址:https://www.cnblogs.com/heyuhhh/p/12051983.html