[多项式ln][多项式exp][多项式求幂][生成函数][DP][FNT] BZOJ 3684: 大朋友和多叉树

SolutionSolution

把DP写成生成函数的形式。f(x)=x+∑d∈Dfd(x)f(x)=x+∑d∈Dfd(x)设g(f(x))=xg(f(x))=x，有g(f(x))g(x)==f(x)−∑d∈Dfd(x)x−∑d∈Dxdg(f(x))=f(x)−∑d∈Dfd(x)g(x)=x−∑d∈Dxd根据拉格朗日反演[xn]h(f(x))=1n[ωn−1]h′(ω)(ωg(ω))n[xn]h(f(x))=1n[ωn−1]h′(ω)(ωg(ω))n令h(x)=xh(x)=x就得到了[xn]f(x)=1n[ωn−1](ωg(ω))n[xn]f(x)=1n[ωn−1](ωg(ω))n现在要求的就是(g(ω)ω)−n(g(ω)ω)−n这个东西。fk(x)=eklnf(x)fk(x)=ekln⁡f(x)lnln直接多项式求逆，expexp牛顿迭代。O(nlogn)O(nlog⁡n)。

#include <bits/stdc++.h>
#define show(x) cerr << #x << " = " << x << endl
using namespace std;
typedef long long ll;
typedef pair<int, int> pairs;

const int N = 303030;
const int MOD = 950009857;

inline char get(void) {
static char buf[100000], *S = buf, *T = buf;
if (S == T) {
T = (S = buf) + fread(buf, 1, 100000, stdin);
if (S == T) return EOF;
}
return *S++;
}
template<typename T>
inline void read(T &x) {
static char c; x = 0; int sgn = 0;
for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
if (sgn) x = -x;
}

inline int pwr(int a, int b) {
int c = 1;
while (b) {
if (b & 1) c = (ll)c * a % MOD;
b >>= 1; a = (ll)a * a % MOD;
}
return c;
}
inline int ivs(int x) {
return pwr(x, MOD - 2);
}
inline int sum(int a, int b) {
a += b;
return a >= MOD ? a - MOD : a;
}
inline int sub(int a, int b) {
return a < b ? a - b + MOD : a - b;
}
inline void add(int &x, int a) {
x = sum(x, a);
}

namespace FNT {
const int MAXN = 303030;
int ww[MAXN], iw[MAXN];
int rev[MAXN];
int num;
inline void pre(int n) {
num = n;
int g = pwr(7, (MOD - 1) / n);
ww[0] = iw[0] = 1;
for (int i = 1; i < num; i++)
iw[n - i] = ww[i] = (ll)ww[i - 1] * g % MOD;
}
inline void fnt(int *a, int n, int f) {
static int x, y, *w;
w = (f == 1) ? ww : iw;
for (int i = 0; i < n; i++)
if (rev[i] > i)
swap(a[rev[i]], a[i]);
for (int i = 1; i < n; i <<= 1)
for (int j = 0; j < n; j += (i << 1))
for (int k = 0; k < i; k++) {
x = a[j + k];
y = (ll)a[j + k + i] * w[num / (i << 1) * k] % MOD;
a[j + k] = sum(x, y);
a[j + k + i] = sub(x, y);
}
if (f == -1){
int in = ivs(n);
for (int i = 0; i < n; i++)
a[i] = (ll)a[i] * in % MOD;
}
}
}

int inv
;

inline void pre(int n) {
inv[1] = 1;
for (int i = 2; i <= n; i++)
inv[i] = (ll)(MOD - MOD / i) * inv[MOD % i] % MOD;
}

void getInv(int *a, int *b, int n) {
using namespace FNT;
static int tmp
;
if (n == 1) return (void)(b[0] = ivs(a[0]));
getInv(a, b, n >> 1);
for (int i = 0; i < n; i++) {
tmp[i] = a[i]; tmp[i + n] = 0;
}
int L = 0; while (!(n >> L & 1)) L++;
for (int i = 0; i < (n << 1); i++)
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << L);
fnt(tmp, n << 1, 1); fnt(b, n << 1, 1);
for (int i = 0; i < (n << 1); i++)
tmp[i] = (ll)b[i] * sub(2, (ll)tmp[i] * b[i] % MOD) % MOD;
fnt(tmp, n << 1, -1);
for (int i = 0; i < n; i++) {
b[i] = tmp[i]; b[n + i] = 0;
}
}
inline void getLn(int *a, int *b, int n) {
using namespace FNT;
static int da
, ia
, tmp
;
for (int i = 0; i < (n << 1); i++)
tmp[i] = da[i] = ia[i] = 0;
getInv(a, ia, n);
for (int i = 1; i < n; i++)
da[i - 1] = (ll)a[i] * i % MOD;
int L = 0; while (!(n >> L & 1)) L++;
for (int i = 0; i < (n << 1); i++)
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << L);
fnt(da, n << 1, 1); fnt(ia, n << 1, 1);
for (int i = 0; i < (n << 1); i++)
tmp[i] = (ll)da[i] * ia[i] % MOD;
fnt(tmp, n << 1, -1);
b[0] = b
= 0;
for (int i = 1; i < n; i++) {
b[i] = (ll)tmp[i - 1] * inv[i] % MOD;
b[n + i] = 0;
}
}
inline void getExp(int *a, int *b, int n) {
using namespace FNT;
static int lb
;
if (n == 1) return (void)(b[0] = 1);
getExp(a, b, n >> 1);
for (int i = 0; i < (n << 1); i++) lb[i] = 0;
getLn(b, lb, n);
for (int i = 0; i < n; i++)
lb[i] = sub(a[i], lb[i]);
lb[0] = sum(lb[0], 1);
int L = 0; while (!(n >> L & 1)) L++;
for (int i = 0; i < (n << 1); i++)
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << L);
fnt(b, n << 1, 1); fnt(lb, n << 1, 1);
for (int i = 0; i < (n << 1); i++)
b[i] = (ll)b[i] * lb[i] % MOD;
fnt(b, n << 1, -1);
for (int i = 0; i < n; i++) b[i + n] = 0;
}
inline void getPow(int *a, int *b, int n, int k) {
static int la
;
k = (k % MOD + MOD) % MOD;
getLn(a, la, n);
for (int i = 0; i < n; i++) la[i] = (ll)la[i] * k % MOD;
getExp(la, b, n);
}

int n, m, x, l;
int g
, f
, g_n
;

int main(void) {
freopen("1.in", "r", stdin);
freopen("1.out", "w", stdout);
FNT::pre(1 << 18);
pre(1 << 18);
read(n); read(m);
for (l = 1; l <= n; l <<= 1);
for (int i = 0; i < m; i++) {
read(x); g[x - 1] = MOD - 1;
}
g[0] = 1;
getPow(g, g_n, l, -n);
cout << (ll)g_n[n - 1] * inv
% MOD << endl;
return 0;
}