【2020省选模拟】题解
T1：
显然莫反
枚举 g c d gcd gcd莫反得
a n s = ∑ d = 1 d ∑ p = 1 n d μ ( p ) f ( n p d ) ans=\sum_{d=1}d\sum_{p=1}^{\frac nd }\mu(p)f(\frac{n}{pd}) ans=d=1∑dp=1∑dnμ(p)f(pdn)
其中 f ( n ) = ∑ i = 1 n ∑ j = 1 i ∑ k = 1 i 1 = n ( n + 1 ) ( 2 n + 1 ) 6 f(n)=\sum_{i=1}^n\sum_{j=1}^i\sum_{k=1}^i1=\frac{n(n+1)(2n+1)}{6} f(n)=∑i=1n∑j=1i∑k=1i1=6n(n+1)(2n+1)
a n s = ∑ T = 1 n g ( T ) f ( n T ) ans=\sum_{T=1}^ng(T)f(\frac n T) ans=T=1∑ng(T)f(Tn)
其中 g ( n ) = ∑ d ∣ n d μ ( n d ) = I d ∗ μ = ϕ g(n)=\sum_{d|n}d\mu(\frac nd)=Id*\mu=\phi g(n)=∑d∣ndμ(dn)=Id∗μ=ϕ
整除分块+杜教筛即可
#includeT2：
l d x ldx ldx日常教育我系列/kk
一个做法是考虑 f ( x ) = ∑ i = 1 x i k f(x)=\sum_{i=1}^xi^k f(x)=∑i=1xik是一个 k + 1 k+1 k+1次多项式
先求出这个多项式，可以插值
那么答案实际就是 ∑ i = 1 n ( n i ) Q i f ( i ) \sum_{i=1}^n{n\choose i}Q^if(i) ∑i=1n(in)Qif(i)
下面用 k k k指代 f ( x ) f(x) f(x)的次数
考虑换一下变成对于每个 t = 0... k t=0...k t=0...k求出 ∑ i = 1 n ( n i ) Q i i t \sum_{i=1}^n{n\choose i}Q^ii^t ∑i=1n(in)Qiit
变一下变成求 n ! Q n ∑ i = 0 n i t i ! Q i − n ( n − i ) ! n!Q^n\sum_{i=0}^n\frac{i^t}{i!}\frac{Q^{i-n}}{(n-i)!} n!Qn∑i=0ni!it(n−i)!Qi−n
设 f t ( x ) = ∑ i = 0 ∞ i t x i i ! f_t(x)=\sum_{i=0}^{\infty}\frac{i^tx^i}{i!} ft(x)=∑i=0∞i!itxi
那么后面就是 [ x n ] ( f t ( x ) e 1 Q x ) [x^n](f_t(x)e^{\frac 1 Qx}) [xn](ft(x)eQ1x)
又有 f t ( x ) = x f t − 1 ′ ( x ) , f 0 = e x f_t(x)=xf'_{t-1}(x),f_0=e^x ft(x)=xft−1′(x),f0=ex
若设 f t ( x ) = e x ∗ A t , 则 A t = x ( A t − 1 + A t − 1 ′ ) f_t(x)=e^x*A_t,则A_t=x(A_{t-1}+A_{t-1}') ft(x)=ex∗At,则At=x(At−1+At−1′)
可以只维护最后 i ∈ [ n − k + 1 , k ] , x i i\in[n-k+1,k],x^i i∈[n−k+1,k],xi这些项即可
复杂度 O ( k 2 ) O(k^2) O(k2)
我的做法
直接将 ∑ i = 0 n ( n i ) Q i ∑ j = 0 i j k \sum_{i=0}^n{n\choose i}Q^i\sum_{j=0}^ij^k ∑i=0n(in)Qi∑j=0ijk后面 j k j^k jk斯特林展开
∑ i = 0 n ( n i ) Q i ∑ j = 0 i ∑ t = 0 k ( j t ) t ! S k , t = ∑ t = 0 k t ! S k , t ∑ i = 0 n ( n i ) Q i ∑ j = 0 i ( j t ) = ∑ t = 0 k t ! S k , t ∑ i = 0 n ( n i ) Q i ( i + 1 t + 1 ) = ∑ t = 0 k t ! S k , t ∑ i = 0 n n ! ( i + 1 ) ! i ! ( n − i ) ! ( t + 1 ) ! ( i − t ) ! Q i = ∑ t = 0 k S k , t t + 1 Q t ∑ i = 0 n − t n ! ( i + t + 1 ) ( n − i − t ) ! i ! Q i = ∑ t = 0 k S k , t Q t n ! ( n − t ) ! ( t + 1 ) ∑ i = 0 n − t ( n − t i ) ( i + t + 1 ) Q i \sum_{i=0}^n{n\choose i}Q^i\sum_{j=0}^i\sum_{t=0}^k{j\choose t}t!S_{k,t}\\ =\sum_{t=0}^kt!S_{k,t}\sum_{i=0}^n{n\choose i}Q^i\sum_{j=0}^{i}{j\choose t}\\ =\sum_{t=0}^kt!S_{k,t}\sum_{i=0}^n{n\choose i}Q^i{i+1\choose t+1}\\ =\sum_{t=0}^kt!S_{k,t}\sum_{i=0}^n\frac{n!(i+1)!}{i!(n-i)!(t+1)!(i-t)!}Q^i\\ =\sum_{t=0}^k\frac{S_{k,t}}{t+1}Q^t\sum_{i=0}^{n-t}\frac{n!(i+t+1)}{(n-i-t)!i!}Q^i\\ =\sum_{t=0}^kS_{k,t}Q^t\frac{n!}{(n-t)!(t+1)}\sum_{i=0}^{n-t}{n-t\choose i}(i+t+1)Q^i i=0∑n(in)Qij=0∑it=0∑k(tj)t!Sk,t=t=0∑kt!Sk,ti=0∑n(in)Qij=0∑i(tj)=t=0∑kt!Sk,ti=0∑n(in)Qi(t+1i+1)=t=0∑kt!Sk,ti=0∑ni!(n−i)!(t+1)!(i−t)!n!(i+1)!Qi=t=0∑kt+1Sk,tQti=0∑n−t(n−i−t)!i!n!(i+t+1)Qi=t=0∑kSk,tQt(n−t)!(t+1)n!i=0∑n−t(in−t)(i+t+1)Qi
然后推到这里的时候傻逼般的卡着了
然后被 l d x ldx ldx大力教育了一顿
对于后面 ∑ i = 0 n − t ( n − t i ) ( i + t + 1 ) Q i = ( t + 1 ) ∑ i = 0 n − t ( n − t i ) Q i + Q ( n − t ) ∑ i = 0 n − t − 1 Q i ( n − t − 1 i ) = ( t + 1 ) ( p + 1 ) n − t + Q ( n − t ) ( Q + 1 ) n − t − 1 \sum_{i=0}^{n-t}{n-t\choose i}(i+t+1)Q^i\\ =(t+1)\sum_{i=0}^{n-t}{n-t\choose i}Q^i+Q(n-t)\sum_{i=0}^{n-t-1}Q^i{n-t-1\choose i}\\ =(t+1)(p+1)^{n-t}+Q(n-t)(Q+1)^{n-t-1} ∑i=0n−t(in−t)(i+t+1)Qi=(t+1)∑i=0n−t(in−t)Qi+Q(n−t)∑i=0n−t−1Qi(in−t−1)=(t+1)(p+1)n−t+Q(n−t)(Q+1)n−t−1
于是后面一整块可以 O ( k ) O(k) O(k)
瓶颈在于前面的斯特林数
复杂度 O ( k 2 / k l o g k ) O(k^2/klogk) O(k2/klogk)
于是可以出到 k = 1 e 6 k=1e6 k=1e6了
我懒，只写了 O ( k 2 ) O(k^2) O(k2)的
注意 k = 0 k=0 k=0的 c o r n e r c a s e corner\ case corner case
#includeT3:
显然期望是积分一下求得
考虑答案即即 ∏ j ( ∑ i ≥ 0 a j i x i ( i + 1 ) ! ) \prod_{j}(\sum_{i\geq0}\frac{a_j^ix^i}{(i+1)!}) j∏(i≥0∑(i+1)!ajixi)
显然要转 ln 后 exp \ln后\exp ln后exp
开始推的是 ln ( e a x − 1 a x ) \ln(\frac{e^{ax}-1}{ax}) ln(axeax−1)，发现这个没法直接求 ln \ln ln
考虑设 f ( x ) = ∑ i x i ( i + 1 ) ! f(x)=\sum_{i}\frac{x^i}{(i+1)!} f(x)=∑i(i+1)!xi
那就是要求 ∑ j ln ( f ( a j x ) ) \sum_j\ln(f(a_jx)) ∑jln(f(ajx))
设 ln ( f ) = ∑ i f i x i \ln(f)=\sum_{i}f_ix^i ln(f)=∑ifixi
那么就是 ∑ j ∑ i = 0 ∞ f i a j i x i = ∑ i f i x i ∑ j = 1 n a j i \sum_j\sum_{i=0}^{\infty}{f_ia_j^ix^i}=\sum_{i}f_ix^i\sum_{j=1}^na_j^i ∑j∑i=0∞fiajixi=∑ifixi∑j=1naji
就是要求等幂和
直接分治 N T T NTT NTT即可
复杂度 O ( n l o g 2 n ) O(nlog^2n) O(nlog2n)
#include