我，理解题解，用了一天
我，卡常数，又用了一天
到了最后，我才发现，我有个加法取模，写的是while(c >= MOD) c -= MOD
我把while改成if，时间，少了
六倍。
六倍。
六倍！！！！
maya我又用第一次T的代码改掉了while，我第一次T的代码也A了= =
那我，改单位复根，FFT循环展开，分治内部循环展开，为了啥= =
好吧，但是我最后上榜了。。。LOJ第四的样子。。
\(\prod_{i = 1}^{N} d_{i}^{m}\sum_{i = 1}^{N}d_{i}^{m}\)
每个联通块的连出去的边，有\(a_{i}\)种可能，所以式子是这样的
\(\prod_{i = 1}^{N} a_{i}^{d_{i}}d_{i}^{m}\sum_{i = 1}^{N}d_{i}^{m}\)
\(\sum_{i = 1}^{N} d_{i}^{m}\prod_{j = 1}^{N}a_{j}^{d_{j}}d_{j}^{m}\)
我们考虑一下一个最暴力的dp（我的第一反应，啥，这怎么是dp？
当然是dp出一个prufer序列啦
\(f[i][j]\)表示考虑了前i个数，填了j个格子，没有多乘任何一个\(d_{i}^{m}\)
\(g[i][j]\)表示考虑了前i个数，填了j个格子，已经乘了一个\(d_{i}^{m}\)
dp的时候，我们只需要枚举下一个数填了多少个格子就好
复杂度\(O(n^{3})\)
但是显然过不掉
我们再考虑一个……套路！因为m出奇的小？
乘方转斯特林数
想一下\(x^{m}\)的组合意义，相当于m种颜色涂在x个格子里，每个种颜色只能涂一个格子，但是每个格子可以涂很多颜色
再看一下原式子，把它变成这样的形式
\(\sum_{i = 1}^{N} a_{i}^{d_{i}}d_{i}^{2m}\prod_{j = 1,j != i}^{N}a_{j}^{d_{j}}d_{j}^{m}\)
然后，我们把乘方拆开，怎么拆呢
考虑到上述的组合意义
我们也就是求所有prufer序列的染色方式，在每一次决策的时候将某个点用2m种颜色染，我们在序列后面（脑补）出1 - N，这样prufer序列里数字出现的个数就正好是点度
设已经被染色的格子是j，那么再放入一个新的数，染色个数为k的时候，需要乘上
\(\binom{n - 2 - j}{k}((S(m,k)k! + S(m,k + 1)(k + 1)!)\)
为什么还有k + 1，因为k代表的是这n - 2个prufer序里面的这类点染色的个数，我们还有后面脑补出的1 - N的位置，所以要+1
同时还要乘上\(a^{k}\)因为同一个位置的点有a种选法
如果是2m种颜色染色，把m换成2m就可以
我们观察一下这个组合数，我们会发现我们可以把\(\frac{1}{k!}\)分离出来，每次统计贡献的时候加上，剩下的等全部算完答案后，如果染色的个数为\(j\)，那么答案再乘上\(\frac{(n - 2)!}{(n - 2 - j)!}\)
同时，我们除了染色的位置还有很多空位，每个空位都有\(\sum_{i = 1}^{n}a_{i}\)种情况，如果有\(j\)个位置被染色了，那么答案还要再乘上
\((\sum_{i = 1}^{n}a_{i})^{n - 2 - j}\)
这样的复杂度是\(n^2m\)的
然而这个方程本质是个卷积，可以上分治FFT
他说不卡常……确实不卡常，我的错误神奇得太离谱了= =
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <map>
//#define ivorysi
#define pb push_back
#define space putchar(‘ ‘)
#define enter putchar(‘\n‘)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define mo 974711
#define MAXN 30005
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < ‘0‘ || c > ‘9‘) {
if(c == ‘-‘) f = -1;
c = getchar();
}
while(c >= ‘0‘ && c <= ‘9‘) {
res = res * 10 + c - ‘0‘;
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {putchar(‘-‘);x = -x;}
if(x >= 10) {
out(x / 10);
}
putchar(‘0‘ + x % 10);
}
const int MOD = 998244353,G = 3,L = (1 << 20);
int fac[MAXN],S[75][75],invfac[MAXN],N,M,a[MAXN],P[MAXN][75],F1[MAXN][75],F2[MAXN][75];
int f[2][MAXN],g[2][MAXN],W[L + 5];
int mul(int a,int b) {return 1LL * a * b % MOD;}
int inc(int a,int b) {a = a + b;if(a >= MOD) a -= MOD;return a;}
void update(int &x,int y) {x = inc(x,y);}
int fpow(int x,int c) {
int res = 1,t = x;
while(c) {
if(c & 1) res = mul(res,t);
t = mul(t,t);
c >>= 1;
}
return res;
}
struct poly {
vector<int> a;
poly() {a.clear();}
friend void NTT(poly &f,int T,int on) {
f.a.resize(T);
for(int i = 1 , j = T / 2; i < T - 1; ++i) {
if(i < j) swap(f.a[i],f.a[j]);
int k = T / 2;
while(j >= k) {j -= k;k >>= 1;}
j += k;
}
for(int h = 2 ; h <= T ; h <<= 1) {
int wn = W[(L + on * L / h) % L];
for(int k = 0 ; k < T ; k += h) {
int w = 1;
for(int j = k ; j < k + h / 2 ; ++j) {
int u = f.a[j],t = mul(f.a[j + h / 2],w);
f.a[j] = inc(u,t);
f.a[j + h / 2] = inc(u,MOD - t);
w = mul(w,wn);
}
}
}
if(on == -1) {
int InvT = fpow(T,MOD - 2);
for(int i = 0 ; i < T ; ++i) f.a[i] = mul(f.a[i],InvT);
}
}
friend poly operator + (const poly &f,const poly &g) {
int T = max(f.a.size(),g.a.size());
poly h;h.a.resize(T);
for(int i = 0 ; i < T ; ++i) h.a[i] = inc(f.a[i],g.a[i]);
return h;
}
friend poly operator * (poly f,poly g) {
int T = 1,t = f.a.size() + g.a.size();
while(T <= t) T <<= 1;
NTT(f,T,1);NTT(g,T,1);
poly h;h.a.resize(T);
for(int i = 0 ; i < T ; ++i) h.a[i] = mul(f.a[i],g.a[i]);
NTT(h,T,-1);
for(int i = T - 1 ; i >= 0 ; --i) {
if(!h.a[i]) h.a.pop_back();
else break;
}
if(h.a.size() > N - 1) h.a.resize(N - 1);
return h;
}
};
void Init() {
read(N);read(M);
int T = max(N,2 * M + 1);
fac[0] = 1;
for(int i = 1 ; i <= T ; ++i) fac[i] = mul(fac[i - 1],i);
invfac[T] = fpow(fac[T],MOD - 2);
for(int i = T - 1; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);
S[0][0] = 1;
for(int i = 1 ; i <= 70 ; ++i) {
for(int j = 1 ; j <= i ; ++j) {
S[i][j] = inc(S[i - 1][j - 1],mul(S[i - 1][j],j));
}
}
for(int i = 1 ; i <= N ; ++i) read(a[i]);
for(int i = 1 ; i <= N ; ++i) {
P[i][0] = 1;
for(int j = 1 ; j <= 70 ; ++j) {
P[i][j] = mul(P[i][j - 1],a[i]);
}
}
W[0] = 1,W[1] = fpow(G,(MOD - 1) / L);
for(int i = 2 ; i < L ; ++i) W[i] = mul(W[i - 1],W[1]);
}
pair<poly,poly> Solve(int L,int R) {
if(L == R) {
poly s,t;s.a.resize(M + 1);t.a.resize(2 * M + 1);
for(int i = 0 ; i <= 2 * M ; ++i) {
if(i <= M)
s.a[i] = mul(mul(inc(mul(S[M][i],fac[i]),mul(S[M][i + 1],fac[i + 1])),invfac[i]),P[L][i]);
t.a[i] = mul(mul(inc(mul(S[2 * M][i],fac[i]),mul(S[2 * M][i + 1],fac[i + 1])),invfac[i]),P[L][i]);
}
return mp(s,t);
}
int mid = (L + R) >> 1;
pair<poly,poly> S = Solve(L,mid),T = Solve(mid + 1,R);
return mp(S.fi * T.fi,S.se * T.fi + S.fi * T.se);
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
pair<poly,poly> res = Solve(1,N);
poly g = res.se;g.a.resize(N - 1);
int sum = 0,p = 1,ans = 0,t = 1;
for(int i = 1 ; i <= N ; ++i) sum = inc(sum,a[i]),p = mul(p,a[i]);
for(int i = N - 2 ; i >= 0; --i) {
ans = inc(ans,mul(mul(g.a[i],t),invfac[N - 2 - i]));
t = mul(t,sum);
}
ans = mul(ans,mul(fac[N - 2],p));
out(ans);putchar(‘\n‘);
return 0;
}#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <map>
//#define ivorysi
#define pb push_back
#define space putchar(‘ ‘)
#define enter putchar(‘\n‘)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define mo 974711
#define MAXN 400005
#define RG register
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < ‘0‘ || c > ‘9‘) {
if(c == ‘-‘) f = -1;
c = getchar();
}
while(c >= ‘0‘ && c <= ‘9‘) {
res = res * 10 + c - ‘0‘;
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {putchar(‘-‘);x = -x;}
if(x >= 10) {
out(x / 10);
}
putchar(‘0‘ + x % 10);
}
const int MOD = 998244353,L = (1 << 16);
int fac[MAXN],S[105][105],invfac[MAXN],a[MAXN],N,M,P[MAXN][105];
int W[L + 5],IW[L + 5],top,fl[MAXN],fr[MAXN],gl[MAXN],gr[MAXN],f[MAXN],g[MAXN];
vector<int> F[105],G[105];
inline int mul(RG const int &a,RG const int &b) {return (int64)a * b % MOD;}
inline int inc(RG const int &a,RG const int &b) {RG int c = a + b;if(c >= MOD) c -= MOD;return c;}
int fpow(RG int x,RG int c) {
RG int res = 1,t = x;
while(c) {
if(c & 1) res = mul(res,t);
t = mul(t,t);
c >>= 1;
}
return res;
}
void NTT(RG int *f,RG int T,RG int on,RG int *w) {
RG int tmp,*wm,*ai,*ami;
for(RG int i = 1 , j = T / 2; i < T - 1 ; ++i) {
if(i < j) {tmp = f[i];f[i] = f[j];f[j] = tmp;}
RG int k = T / 2;
while(j >= k) {
j -= k;
k >>= 1;
}
j += k;
}
#define work(j) {tmp = mul(ami[j],wm[j]);ami[j] = inc(ai[j],MOD - tmp);ai[j] = inc(ai[j],tmp); }
for(RG int h = 2,m = 1; h <= T ; m = h,h <<= 1) {
wm = w + m;
if(m < 8) {
for(RG int i = 0 ; i < T ; i += h) {
ai = f + i,ami = f + m + i;
for(RG int j = 0 ; j < m ; ++j) work(j);
}
}
else {
for(RG int i = 0 ; i < T ; i += h) {
ai = f + i,ami = f + m + i;
for(RG int j = 0 ; j < m ; j += 8) {
work(j);
work(j + 1);
work(j + 2);
work(j + 3);
work(j + 4);
work(j + 5);
work(j + 6);
work(j + 7);
}
}
}
}
if(on < 0) {
RG int InvT = fpow(T,MOD - 2);
#define C(x,y) {f[x] = mul(f[x],y);}
if(T < 8) {for(RG int i = 0 ; i < T ; ++i) C(i,InvT);}
else {
for(RG int i = 0 ; i < T ; i += 8) {
C(i,InvT);
C(i + 1,InvT);
C(i + 2,InvT);
C(i + 3,InvT);
C(i + 4,InvT);
C(i + 5,InvT);
C(i + 6,InvT);
C(i + 7,InvT);
}
}
}
}
void Init() {
read(N);read(M);
RG int T = max(N,2 * M + 1);
fac[0] = 1;
for(RG int i = 1 ; i <= T ; ++i) fac[i] = mul(fac[i - 1],i);
invfac[T] = fpow(fac[T],MOD - 2);
for(RG int i = T - 1; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);
S[0][0] = 1;
for(RG int i = 1 ; i <= 70 ; ++i) {
for(RG int j = 1 ; j <= i ; ++j) {
S[i][j] = inc(S[i - 1][j - 1],mul(S[i - 1][j],j));
}
}
for(RG int j = 1 ; j <= 2 * M ; ++j) {
S[M][j] = mul(S[M][j],fac[j]);
S[2 * M][j] = mul(S[2 * M][j],fac[j]);
}
for(RG int j = 0 ; j <= 2 * M ; ++j) {
S[M][j] = mul(inc(S[M][j],S[M][j + 1]),invfac[j]);
S[2 * M][j] = mul(inc(S[2 * M][j],S[2 * M][j + 1]),invfac[j]);
}
for(RG int i = 1 ; i <= N ; ++i) read(a[i]);
for(RG int i = 1 ; i <= N ; ++i) {
P[i][0] = 1;
for(RG int j = 1 ; j <= 70 ; ++j) {
P[i][j] = mul(P[i][j - 1],a[i]);
}
}
RG int half = L / 2;
RG int t1 = fpow(3,(MOD - 1) / L),t2 = fpow(t1,MOD - 2);
W[half] = 1;IW[half] = 1;
for(RG int i = 1 ; i < half; ++i) W[i + half] = mul(W[i + half - 1],t1),IW[i + half] = mul(IW[i + half - 1],t2);
for(RG int i = half - 1 ; i >= 0 ; --i) W[i] = W[i << 1],IW[i] = IW[i << 1];
}
void Solve(RG int L,RG int R) {
if(L == R) {
++top;
F[top].resize(M + 1);G[top].resize(2 * M + 1);
for(RG int i = 0 ; i <= 2 * M ; ++i) {
if(i <= M)
F[top][i] = mul(S[M][i],P[L][i]);
G[top][i] = mul(S[2 * M][i],P[L][i]);
}
return ;
}
RG int mid = (L + R) >> 1;
Solve(L,mid);int Ld = top;
Solve(mid + 1,R);int Rd = top;
top -= 2;
RG int s1 = F[Ld].size(),s2 = F[Rd].size(),s3 = G[Ld].size(),s4 = G[Rd].size();
RG int t = max(max(s1 + s2,s1 + s4),s3 + s2);
RG int K = 1;while(K <= t) K <<= 1;
for(int i = 0 ; i < s1 ; ++i) fl[i] = F[Ld][i];
for(int i = 0 ; i < s2 ; ++i) fr[i] = F[Rd][i];
for(int i = 0 ; i < s3 ; ++i) gl[i] = G[Ld][i];
for(int i = 0 ; i < s4 ; ++i) gr[i] = G[Rd][i];
fill(fl + s1,fl + K,0);
fill(fr + s2,fr + K,0);
fill(gl + s3,gl + K,0);
fill(gr + s4,gr + K,0);
NTT(fl,K,1,W);NTT(fr,K,1,W);NTT(gl,K,1,W);NTT(gr,K,1,W);
#define Calc1(i) {f[i] = mul(fl[i],fr[i]);}
#define Calc2(i) {g[i] = inc(mul(fl[i],gr[i]),mul(gl[i],fr[i]));}
if(K < 8) {
for(int i = 0 ; i < K ; ++i) {
Calc1(i);Calc2(i);
}
}
else {
for(RG int i = 0 ; i < K ; i += 8) {
Calc1(i);Calc1(i + 1);
Calc1(i + 2);Calc1(i + 3);
Calc1(i + 4);Calc1(i + 5);
Calc1(i + 6);Calc1(i + 7);
Calc2(i);Calc2(i + 1);
Calc2(i + 2);Calc2(i + 3);
Calc2(i + 4);Calc2(i + 5);
Calc2(i + 6);Calc2(i + 7);
}
}
NTT(f,K,-1,IW);NTT(g,K,-1,IW);
++top;
F[top].clear();G[top].clear();
t = min(N - 2,K - 1);
while(t >= 0) {
if(!f[t]) --t;
else break;
}
for(RG int i = 0 ; i <= t ; ++i) F[top].pb(f[i]);
t = min(N - 2,K - 1);
while(t >= 0) {
if(!g[t]) --t;
else break;
}
for(RG int i = 0 ; i <= t ; ++i) G[top].pb(g[i]);
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve(1,N);
G[top].resize(N - 1);
RG int ans = 0,sum = 0,p = 1,t = 1;
for(RG int i = 1 ; i <= N ; ++i) sum = inc(sum,a[i]),p = mul(p,a[i]);
for(RG int i = N - 2 ; i >= 0 ; --i) {
ans = inc(ans,mul(mul(G[top][i],t),invfac[N - 2 - i]));
t = mul(t,sum);
}
ans = mul(mul(ans,p),fac[N - 2]);
out(ans);enter;
//out(clock());enter;
return 0;
}原文：https://www.cnblogs.com/ivorysi/p/9093086.html