UOJ269 如何优雅地求和
Problem
UOJ
给定 n , p n,p n,p ， f ( x ) f(x) f(x)是一个 m m m 阶函数，求
(1) Q ( f ) = ∑ k = 0 n f ( k ) ( n k ) p k ( 1 − p ) n − k Q(f)=\sum_{k=0}^n f(k)\binom n k p^k (1-p)^{n-k}\tag1 Q(f)=k=0∑nf(k)(kn)pk(1−p)n−k(1)
Solution
首先 ( n − m ) = 0 \binom {n} {-m}=0 (−mn)=0
我们思考简单的二项分布即当 f ( x ) = x f(x)=x f(x)=x时的推导过程：
(2) Q ( f ) = ∑ k = 0 n k n ! k ! ( n − k ) ! p k ( 1 − p ) n − k Q(f)=\sum_{k=0}^n k \frac {n!} {k!(n-k)!} p^k (1-p)^{n-k}\tag2 Q(f)=k=0∑nkk!(n−k)!n!pk(1−p)n−k(2)
(3) Q ( f ) = ∑ k = 0 n n ( n − 1 ) ! ( k − 1 ) ! ( n − k ) ! p k ( 1 − p ) n − k Q(f)=\sum_{k=0}^n n\frac {(n-1)!} {(k-1)!(n-k)!} p^k (1-p)^{n-k}\tag3 Q(f)=k=0∑nn(k−1)!(n−k)!(n−1)!pk(1−p)n−k(3)
(4) Q ( f ) = n p ∑ k = 0 n ( n − 1 k − 1 ) p k − 1 ( 1 − p ) n − k Q(f)=np\sum_{k=0}^n \binom {n-1} {k-1} p^{k-1} (1-p)^{n-k}\tag4 Q(f)=npk=0∑n(k−1n−1)pk−1(1−p)n−k(4)
(5) Q ( f ) = n p ∑ k = 0 n − 1 ( n − 1 k ) p k ( 1 − p ) n − k − 1 Q(f)=np\sum_{k=0}^{n-1} \binom {n-1} {k} p^k (1-p)^{n-k-1}\tag5 Q(f)=npk=0∑n−1(kn−1)pk(1−p)n−k−1(5)
(6) Q ( f ) = n p ( p + ( 1 − p ) ) n − 1 = n p Q(f)=np(p+(1-p))^{n-1}=np\tag6 Q(f)=np(p+(1−p))n−1=np(6)
如法炮制，当 f ( x ) = x d ‾ f(x)=x^{\underline d} f(x)=xd时
(7) Q ( f ) = n d ‾ p d ∑ k = 0 n − d ( n − d k ) p k ( 1 − p ) n − k = n d ‾ p d Q(f)=n^{\underline d}p^d\sum_{k=0}^{n-d} \binom {n-d} {k}p^{k}(1-p)^{n-k}=n^{\underline d}p^d\tag7 Q(f)=ndpdk=0∑n−d(kn−d)pk(1−p)n−k=ndpd(7)
那么我们希望能求出 b i b_i bi使得 f ( x ) = ∑ i = 0 m b i x i ‾ f(x)=\sum_{i=0}^m b_i x^{\underline i} f(x)=∑i=0mbixi，这样 Q ( f ) = ∑ i = 0 m b i n i ‾ p i Q(f)=\sum_{i=0}^mb_in^{\underline i}p^i Q(f)=∑i=0mbinipi
不妨令 c i = i ! b i c_i=i!b_i ci=i!bi，那么
(8) f ( x ) = ∑ i = 0 m c i x i ‾ i ! = ∑ i = 0 m c i ( x i ) f(x)=\sum_{i=0}^m c_i\frac {x^{\underline i}} {i!}=\sum_{i=0}^m c_i \binom {x} {i}\tag8 f(x)=i=0∑mcii!xi=i=0∑mci(ix)(8)
(9) Δ f ( x ) = f ( x + 1 ) − f ( x ) = ∑ i = 0 m c i ( ( x + 1 i ) − ( x i ) ) = ∑ i = 0 m c i ( x i − 1 ) \Delta f(x)=f(x+1)-f(x)=\sum_{i=0}^m c_i\biggl(\binom {x+1} {i}-\binom {x} {i}\biggr)=\sum_{i=0}^{m} c_i\binom {x} {i-1}\tag9 Δf(x)=f(x+1)−f(x)=i=0∑mci((ix+1)−(ix))=i=0∑mci(i−1x)(9)
取 x = 0 x=0 x=0则仅有一个组合数为1，即 Δ f ( 0 ) = c 1 \Delta f(0)=c_1 Δf(0)=c1。
然后一阶差分的式子和原来的式子惊人地相似，那么我们可以继续利用差分推出 Δ k f ( 0 ) = c k \Delta^kf(0)=c_k Δkf(0)=ck
暴力差分的时间复杂度是 O ( m 2 ) O(m^2) O(m2) ，这样就可以过了。但其实 k k k 阶差分是可以继续优化的，通过找规律，我们会发现
(10) Δ k f ( 0 ) = ∑ i = 0 k ( − 1 ) k − i ( k i ) a i \Delta^kf(0)=\sum_{i=0}^k (-1)^{k-i}\binom k i a_i\tag{10} Δkf(0)=i=0∑k(−1)k−i(ik)ai(10)
NTT加速即可做到 O ( m log m ) O(m\log m) O(mlogm)。
Code
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