多项式全家桶

对 $ln$ 的处理是，先求导再积分。(详见付公主的背包)

当我们在对分式求导时，可以把他先化为母函数的无穷级数，然后求导，再化回来。

[ZJOI2019] 开关

不愧是浙江省选，难到让我失去写题解的欲望了。有需求可以看粉兔题解。

[WC2019] 数树

问题0

只用考虑边的交集大小。

问题1

引理： 子集容斥： $f(S) = \sum_{T \sube S} \sum_{Q \sube T} (-1)^{|T| - |Q|} f(Q)$

则答案为：

\sum_{E 2} y^{n - | E 1 \land E 2 |} \sum_{T \subseteq E 2} g (T) \sum_{Q \subseteq T} (- 1)^{| T | - | Q |} y^{n - | Q |} \sum_{T \subseteq E 2} g (T) y^{n - | T |} \sum_{Q \subseteq T} (- y)^{| T | - | Q |} \sum_{T \subseteq E 2} g (T) y^{n - | T |} (1 - y)^{| T |} \sum_{T \subseteq E 2} y^{k} (1 - y)^{n - k} n^{k - 2} \prod_{i = 1}^{k} a_{i} \frac{(1 - y)^{n}}{n^{2}} \sum_{T \subseteq E 2} \prod_{i = 1}^{k} \frac{n y}{1 - y} a_{i}

$\sum_{E2} y^{n - |E1 \and E2|} \\ \sum_{T \sube E2} g(T) \sum_{Q \sube T} (-1)^{|T| - |Q|} y^{n - |Q|} \\ \sum_{T \sube E2} g(T) y^{n - |T|} \sum_{Q \sube T} (-y)^{|T| - |Q|} \\ \sum_{T \sube E2} g(T) y^{n - |T|} (1 - y)^{|T|} \\ \sum_{T \sube E2} y^{k} (1 - y)^{n - k} n^{k-2} \prod_{i=1}^k a_i \\ \frac{(1-y)^n}{n^2} \sum_{T \sube E2} \prod_{i=1}^k \frac{ny}{1 - y} a_i$

问题2

先书写最暴力的式子。

\sum_{E 1} \sum_{E 2} y^{n - | E 1 \land E 2 |} \sum_{T} g^{2} (T) y^{n - | T |} \sum_{Q \subseteq T} (- y)^{| T | - | Q |} \sum_{T} g^{2} (T) y^{n - | T |} (1 - y)^{| T |} \sum_{T} y^{k} (1 - y)^{n - k} n^{2 k - 4} \prod_{i = 1}^{k} a_{i}^{2} \frac{(1 - y)^{n}}{n^{4}} \sum_{T} \prod_{i = 1}^{k} \frac{n^{2} y}{(1 - y)} a_{i}^{2} F (x) = \sum_{i = 1} \frac{n^{2} y}{(1 - y) i!} i^{i} x^{i} \sum_{T} \prod_{i = 1}^{k} \frac{n^{2} y}{(1 - y)} a_{i}^{2} = n! [x^{n}] \sum_{k = 0} \frac{1}{k!} F^{k} (x) = n! [x^{n}] e^{F (x)}

$\sum_{E1}\sum_{E2} y^{n - |E1 \and E2|} \\ \sum_{T} g^2(T)y^{n - |T|} \sum_{Q \sube T} (-y)^{|T| - |Q|} \\ \sum_{T} g^2(T)y^{n - |T|} (1 - y)^{|T|} \\ \sum_{T} y^{k}(1 - y)^{n - k} n^{2k - 4} \prod_{i = 1}^{k} a_i^2 \\ \frac{(1-y)^n}{n^4} \sum_{T} \prod_{i=1}^k \frac{n^2y}{(1-y)} a_i^2 \\ F(x) = \sum_{i=1} \frac{n^2y}{(1-y)i!} i^i x^i \\ \sum_{T} \prod_{i=1}^k \frac{n^2y}{(1-y)} a_i^2 = n! [x^n] \sum_{k=0} \frac{1}{k!} F^k(x) = n! [x^n]e^{F(x)}$

[AGC038E] Gachapon

类似开关，也是高妙生成函数题（但好像可以minmax容斥）。

我们定义 $f_i$ 表示在第 $i$ 轮停止的概率。

我们钦定第一个最后被选，则他的指数型生成函数为：

x \frac{1}{(B_{1} - 1)!} p_{1}^{B_{1}} x^{B_{1} - 1} \prod_{i = 2}^{n} (e^{p_{i} x} - \sum_{j = 0}^{B_{i} - 1} \frac{1}{j!} p_{i}^{j} x^{j})

$x\frac{1}{(B_1-1)!}p_1^{B_1}x^{B_{1} - 1} \prod_{i=2}^n (e^{p_ix} - \sum_{j=0}^{B_i-1} \frac{1}{j!} p_i^jx^j)$

显然可以背包展开系数为 $a_{w,t} x^{w} e^{tx}$ 的形式,整理可得：

f_{k + 1} = \sum a_{w, t} t^{k - w} \frac{k!}{(k - w)!}

$f_{k+1} = \sum a_{w , t} t^{k-w} \frac{k!}{(k-w)!} \\$

不妨提出化为组合数的形式。

\sum (w + 1)! a_{w, t} \sum t^{k - w} (\binom{k + 1}{k - w}) = \sum (w + 1)! a_{w, t} \frac{1}{(1 - t)^{w + 2}}

$\sum (w + 1)! a_{w,t} \sum t^{k-w} \binom{k + 1}{k - w} \\ = \sum (w + 1)! a_{w,t} \frac{1}{(1 - t)^{w + 2}}$

最后得到以上式子，背包的部分我们把钦定的东西拿来一起背包，同时记录以下是否完成钦定就好了。

#include <set>
#include <map>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define pii pair <int , int>
#define mp make_pair
#define fs first
#define sc second
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;

//const int Mxdt=100000; 
//static char buf[Mxdt],*p1=buf,*p2=buf;
//#define getchar() p1==p2&&(p2=(p1=buf)+fread(buf,1,Mxdt,stdin),p1==p2)?EOF:*p1++;
template <typename T>
void read(T &x) {
	x=0;T f=1;char s=getchar();
	while(s<'0'||s>'9') {if(s=='-') f=-1;s=getchar();}
	while(s>='0'&&s<='9') {x=(x<<3)+(x<<1)+(s^'0');s=getchar();}
	x *= f;
}

template <typename T>
void write(T x , char s='\n') {
	if(x<0) {putchar('-');x=-x;}
	if(!x) {putchar('0');putchar(s);return;}
	T tmp[25] = {} , t = 0;
	while(x) tmp[t ++] = x % 10 , x /= 10;
	while(t -- > 0) putchar(tmp[t] + '0');
	putchar(s);
}

const int MAXN = 405;
const int mod = 998244353;

inline int Add(int x , int y) {x += y;return x >= mod?x - mod:x;}
inline int Sub(int x , int y) {x -= y;return x < 0?x + mod:x;}
inline int Mul(int x , int y) {return 1ll * x * y % mod;}

int qpow(int a , int b) {
    int res = 1;
    while(b) {
        if(b & 1) res = Mul(res , a);
        a = Mul(a , a);
        b >>= 1;
    }
    return res;
}

int fac[MAXN] , finv[MAXN] , n , a[MAXN] , b[MAXN];
int p[MAXN][MAXN] , dp[2][MAXN][MAXN][2] , A , B , PA;

int main() {
    read(n);
    for (int i = 1; i <= n; ++i) read(a[i]),read(b[i]) , A = Add(A , a[i]) , B = Add(B , b[i]);
    fac[0] = 1;
    int N = 400;
    for (int i = 1; i <= N + 1; ++i) fac[i] = Mul(fac[i - 1] , i);
    finv[N + 1] = qpow(fac[N + 1] , mod - 2);
    for (int i = N + 1; i >= 1; --i) finv[i - 1] = Mul(finv[i] , i);
    PA = qpow(A , mod - 2);
    for (int i = 1; i <= n; ++i) {
        p[i][0] = 1 , p[i][1] = Mul(a[i] , PA);
        for (int j = 2; j <= b[i]; ++j) p[i][j] = Mul(p[i][j - 1] , p[i][1]);
    }
    dp[0][0][0][0] = 1;
    for (int i = 1 , preA = a[1] , preB = b[1]; i <= n; ++i , preA += a[i] , preB += b[i]) {
        for (int j = preA; j >= 0; --j) for (int k = preB; k >= 0; --k) {
            dp[i & 1][j][k][0] = dp[i & 1][j][k][1] = 0;
            int op = (i - 1) & 1;
            if(j >= a[i]) {
                dp[i & 1][j][k][0] = Add(dp[op][j - a[i]][k][0] , dp[i & 1][j][k][0]);
                dp[i & 1][j][k][1] = Add(dp[op][j - a[i]][k][1] , dp[i & 1][j][k][1]);
            }
            for (int w = 0; w < b[i] && w <= k; ++w) {
                dp[i & 1][j][k][0] = Add(Mul(dp[op][j][k - w][0] , Mul(Mul(mod - 1 , finv[w]) , p[i][w])) , dp[i & 1][j][k][0]);
                dp[i & 1][j][k][1] = Add(Mul(dp[op][j][k - w][1] , Mul(Mul(mod - 1 , finv[w]) , p[i][w])) , dp[i & 1][j][k][1]);
            }
            if(k >= b[i] - 1) dp[i & 1][j][k][1] = Add(dp[i & 1][j][k][1] , Mul(dp[op][j][k - (b[i] - 1)][0] , Mul(p[i][b[i]] , finv[b[i] - 1])));
        }
    }
    int ans = 0;
    for (int t = 0; t < A; ++t) for (int w = 0; w <= B; ++w) {
        ans = Add(ans , Mul(fac[w + 1] , Mul(dp[n & 1][t][w][1] , qpow(qpow(Sub(1 , Mul(t , PA)) , w + 2) , mod - 2))));
    }
    write(ans);
    return 0;
}

相关知识以后补，先存份代码。

#include <set>
#include <map>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define pii pair <int , int>
#define mp make_pair
#define fs first
#define sc second
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;


//const int Mxdt=100000; 
//static char buf[Mxdt],*p1=buf,*p2=buf;
//#define getchar() p1==p2&&(p2=(p1=buf)+fread(buf,1,Mxdt,stdin),p1==p2)?EOF:*p1++;
template <typename T>
void read(T &x) {
	x=0;T f=1;char s=getchar();
	while(s<'0'||s>'9') {if(s=='-') f=-1;s=getchar();}
	while(s>='0'&&s<='9') {x=(x<<3)+(x<<1)+(s^'0');s=getchar();}
	x *= f;
}

template <typename T>
void write(T x , char s='\n') {
	if(x<0) {putchar('-');x=-x;}
	if(!x) {putchar('0');putchar(s);return;}
	T tmp[25] = {} , t = 0;
	while(x) tmp[t ++] = x % 10 , x /= 10;
	while(t -- > 0) putchar(tmp[t] + '0');
	putchar(s);
}

const int MAXN = 2e6 + 5;
const int mod = 998244353;

inline int Add(int x , int y) {x += y;return x >= mod?x - mod:x;}
inline int Sub(int x , int y) {x -= y;return x < 0?x + mod:x;}
inline int Mul(int x , int y) {return 1ll * x * y % mod;}

inline int qpow(int a , int b) {
	int res = 1;
	while(b) {
		if(b & 1) res = Mul(res , a);
		a = Mul(a , a);
		b >>= 1;
	}
	return res;
}


namespace Poly_Family {
        
        inline int Cipolla(int n , int mod) {
        	if(!n) return 0;
        	if(qpow(n , (mod - 1) / 2) != 1) return -1;
        	int a = 1;
        	while(1) {
        		a = rand() % mod;
        		if(a && qpow(Sub(Mul(a , a) , n) , (mod - 1) / 2) != 1) break; 
        	}
        	const int I_P = Sub(Mul(a , a) , n);
        	pii A = mp(a , 1) , res = mp(1 , 0);
        	int b = (mod + 1) / 2;
        	while(b) {
        		if(b & 1) {
        			pii tmp;
        			tmp.fs = Add(Mul(res.fs , A.fs) , Mul(I_P , Mul(res.sc , A.sc)));
        			tmp.sc = Add(Mul(res.fs , A.sc) , Mul(res.sc , A.fs));
        			res = tmp;
        		}
        		pii tmp;
        		tmp.fs = Add(Mul(A.fs , A.fs) , Mul(I_P , Mul(A.sc , A.sc)));
        		tmp.sc = Add(Mul(A.fs , A.sc) , Mul(A.sc , A.fs));
        		A = tmp;
        		b >>= 1;
        	}
        	int x = res.fs , y = Sub(0 , x); 
        	return min(x , y);
        }
	#define poly vector <int> 
	#define Len(x) (int)x.size()
	
	int r[MAXN << 2];
	poly w[2][23];
	
	inline void pre(int op) {
		for (int t = 1; t <= 22; ++t) {
			w[op][t].resize(1 << t);
			int k = (1 << (t - 1));
			int ori = qpow(3 , (op == 1)?(mod - 1) / (k << 1):(mod - 1) - (mod - 1) / (k << 1));
			int now = 1;
			for (int j = 0; j < k; ++j) {
				w[op][t][j] = now;
				now = Mul(now , ori);
			}
		}
	}
	
	inline void Resize(poly &F , int n) {F.resize(n , 0);}
	
	inline void NTT(poly &F , int op) {
		if(op == -1) op = 0;
		int N = Len(F);
		for (int i = 0; i < N; ++i) if(i < r[i]) swap(F[i] , F[r[i]]);
		for (int k = 1 , t = 1; k < N; k <<= 1 , t ++) {
			for (int i = 0; i < (N >> t); ++i) {
				for (int j = 0; j < k; ++j) {
					int cur = Mul(w[op][t][j] , F[(i << t) ^ k ^ j]);
					F[(i << t) ^ j ^ k] = Sub(F[(i << t) ^ j] , cur);
					F[(i << t) ^ j] = Add(F[(i << t) ^ j] , cur);
				}
			}
		} 
		if(op == 0) {
			int D = qpow(N , mod - 2);
			for (int i = 0; i < N; ++i) F[i] = Mul(F[i] , D);
		}
	}
	
	inline poly operator - (poly F , poly G) {
		int n = Len(F) , m = Len(G) , N = max(n , m);
		Resize(F , N) , Resize(G , N);
		for (int i = 0; i < Len(F); ++i) F[i] = Sub(F[i] , G[i]) % mod;
		return F;
	}

	inline poly operator + (poly F , poly G) {
		int n = Len(F) , m = Len(G) , N = max(n , m);
		Resize(F , N) , Resize(G , N);
		for (int i = 0; i < Len(F); ++i) F[i] = Add(F[i] , G[i]) % mod;
		return F;
	}
	
	inline poly operator * (poly F , poly G) {
		poly FG;
		int n = Len(F) , m = Len(G) , rl = n + m - 1 , N = 1;
		while(N < rl) N <<= 1;
		Resize(F , N) , Resize(G , N) , Resize(FG , N);
		for (int i = 1; i < N; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1)?(N >> 1):0);
		NTT(F , 1) , NTT(G , 1);
		for (int i = 0; i < N; ++i) FG[i] = Mul(F[i] , G[i]);
		NTT(FG , -1);
		Resize(FG , rl);
		return FG;
	}
	
	inline poly Deri(poly F) {
		for (int i = 0; i < Len(F) - 1; ++i) F[i] = Mul(F[i + 1] , i + 1);
		F[Len(F) - 1] = 0;
		return F;
	}
	
	inline poly Integ(poly F) {
		Resize(F , Len(F) + 1);
		for (int i = Len(F) - 1; i >= 1; --i) F[i] = Mul(F[i - 1] , qpow(i , mod - 2));
		F[0] = 0;
		return F;
	}
	
	inline poly Inv(poly F) {
		poly G;
		G.resize(1);
		G[0] = qpow(F[0] , mod - 2);
		for (int k = 2; k <= 2 * Len(F); k <<= 1) {
			poly A = F;
			Resize(A , k);
			
			Resize(G , k);
			for (int i = 1; i < k; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1)?(k >> 1):0);
			
			NTT(G , 1);
			poly tmp;Resize(tmp , k);
			for (int i = 0; i < Len(G); ++i) tmp[i] = Mul(G[i] , G[i]);
			NTT(tmp , -1) , NTT(G , -1);
			tmp = tmp * A;
			Resize(tmp , k);
			for (int i = 0; i < Len(G); ++i) G[i] = Mul(G[i] , 2);
			G = G - tmp;
		}
		return G;
	}
	
	inline poly operator / (poly F , poly G) {
		reverse(F.begin() , F.end());
		reverse(G.begin() , G.end());
		int n = Len(F) , m = Len(G);
		int N = n - m + 1;
		Resize(G , N);
		G = Inv(G);
		poly Q = F * G;Resize(Q , N);
		reverse(Q.begin() , Q.end());
		return Q;
	}
	
	inline poly Ln(poly F) {
		poly G = Integ(Inv(F) * Deri(F));
		return G;
	}
	
	inline poly Exp(poly F) {
		poly f;Resize(f , 1);
		f[0] = 1;
		for (int k = 2; k <= 2 * Len(F); k <<= 1) {
			poly A = F;
			Resize(A , k);
			A = A - Ln(f);
			A[0] = Add(A[0] , 1);
			f = f * A;
			Resize(f , k);
		}
		return f;
	}
	
	inline poly Qpow(poly F , int k1 , int k2 , int p) {
		int l = -1;
		for (int i = 0; i < Len(F); ++i) if(F[i]) {
			l = i;
			break;
		}
		if(l == -1) return F;
		if((LL)l * k1 + p * mod * (l > 0) >= Len(F)) {
			for (int i = 0; i < Len(F); ++i) F[i] = 0;
			return F;
		}
		poly G;G.resize(Len(F) - l);
		for (int i = 0; i < Len(G); ++i) G[i] = F[i + l];
		int v = G[0] , iv = qpow(v , mod - 2);
		for (int i = 0; i < Len(G); ++i) G[i] = Mul(G[i] , iv); 
		
		G = Ln(G),Resize(G , Len(F) - l);
		for (int i = 0; i < Len(G); ++i) G[i] = Mul(G[i] , k1);
		G = Exp(G),Resize(G , Len(F) - l);
		
		v = qpow(v , k2);
		for (int i = 0; i < Len(G); ++i) G[i] = Mul(G[i] , v);
		
		l *= k1;
		for (int i = 0; i < l; ++i) F[i] = 0;
		for (int i = l; i < Len(F); ++i) F[i] = G[i - l]; 
		
		return F;
	}
	
	inline poly Sqrt(poly F) {
		poly f;Resize(f , 1);
		f[0] = Cipolla(F[0] , mod);
		if(f[0] == -1) return f;
		int inv2 = (mod + 1) >> 1;
		for (int k = 2; k <= 2 * Len(F); k <<= 1) {
			poly A = F;
			Resize(A , k);
			poly tmp = Inv(f);
			Resize(tmp , k);
			f = f + tmp * A;
			Resize(f , k);
			for (int i = 0; i < Len(f); ++i) f[i] = Mul(f[i] , inv2);
			Resize(f , k);
		}
		return f;
	}
}

using namespace Poly_Family;

poly F;
int n;

int main() {
	pre(0) , pre(1);
	int n , k1 = 0 , k2 = 0 , p = 0;
	read(n);
	
	char s=getchar();
	while(s<'0'||s>'9') s=getchar();
	while(s>='0'&&s<='9') {p |= (k1 * 10ll + (s ^ '0') >= mod) , k1 = Add(Mul(k1 , 10) , s ^ '0') , k2 = (k2 * 10ll % (mod - 1) + (s ^ '0')) % (mod - 1);s=getchar();}
	
	Resize(F , n);
	for (int i = 0; i < n; ++i) read(F[i]);
	
	F = Qpow(F , k1 , k2 , p);
	
	for (int i = 0; i < Len(F); ++i) write(F[i] , (i!=Len(F))?' ':'\n');
	return 0;
}

posted @ 2022-05-06 21:17 Reanap 阅读(51) 评论(0) 编辑收藏举报

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