牛客竞赛数学专题班生成函数I 题解

题单链接

背包

题目链接

题意

总共有888个物品，对于每个物品的选法都有要求，问带nnn个物品的方案数。

思路

构造生成函数，并将等比级数转为合式(∏i=0xi=11−x\prod_{i=0}x^i=\frac{1}{1-x}∏i=0xi=1−x1)：

肥宅快乐水最多1个: 1+x1+x1+x
大盘鸡最多2个: 1+x+x2=11−x−x31−x=1−x31−x1+x+x^2=\frac{1}{1-x}-\frac{x^3}{1-x}=\frac{1-x^3}{1-x}1+x+x2=1−x1−1−xx3=1−x1−x3
啤酒鸡最多带3个: 1+x+x2+x3=11−x−x41−x=1−x41−x1+x+x^2+x^3=\frac{1}{1-x}-\frac{x^4}{1-x}=\frac{1-x^4}{1-x}1+x+x2+x3=1−x1−1−xx4=1−x1−x4
鸡翅的个数是2的倍数: 1+x2+x4+x6...=11−x21+x^2+x^4+x^6...=\frac{1}{1-x^2}1+x2+x4+x6...=1−x21
鸡汤的个数是奇数: x1+x3+x5...=x1−x2x^1+x^3+x^5...=\frac{x}{1-x^2}x1+x3+x5...=1−x2x
鸡块的个数是4的倍数: 1+x4+x8+x12...=11−x41+x^4+x^8+x^{12}...=\frac{1}{1-x^4}1+x4+x8+x12...=1−x41
鸡腿最多带一个: 1+x1+x1+x
鸡蛋的个数是3的倍数: 1+x3+x6+x9+...=11−x31+x^3+x^6+x^9+...=\frac{1}{1-x^3}1+x3+x6+x9+...=1−x31

将上述的式子全部乘起来（卷积）：
F(x)=(1+x)⋅(1−x31−x)⋅(1−x41−x)⋅(11−x2)⋅(x1−x2)⋅(11−x4)⋅(1+x)⋅11−x3=x1−x4F(x)=(1+x)\cdot (\frac{1-x^3}{1-x}) \cdot (\frac{1-x^4}{1-x}) \cdot (\frac{1}{1-x^2}) \cdot (\frac{x}{1-x^2}) \cdot (\frac{1}{1-x^4}) \cdot (1+x)\cdot \frac{1}{1-x^3} = \frac{x}{1-x^4} F(x)=(1+x)⋅(1−x1−x3)⋅(1−x1−x4)⋅(1−x21)⋅(1−x2x)⋅(1−x41)⋅(1+x)⋅1−x31=1−x4x
由公式A(x)=∑i≥0(i+k−1i)xi=1(1−x)kA(x)=\sum_{i\ge 0}\binom{i+k-1}{i}x^i=\frac{1}{(1-x)^k}A(x)=∑i≥0(ii+k−1)xi=(1−x)k1，可得:
F(x)=x1−x4=x⋅∑i≥0(i+4−1i)xi=x⋅∑i≥0(i+33)xi=∑n≥1(n+23)xnF(x)=\frac{x}{1-x^4}=x\cdot \sum_{i\ge 0}\binom{i+4-1}{i}x^i=x\cdot \sum_{i\ge 0}\binom{i+3}{3}x^i=\sum_{n\ge 1}\binom{n+2}{3}x^n F(x)=1−x4x=x⋅i≥0∑(ii+4−1)xi=x⋅i≥0∑(3i+3)xi=n≥1∑(3n+2)xn
答案为：
[xn]F(x)=(n+23)=(n+2)(n+1)n3×2[x^n]F(x)=\binom{n+2}{3}=\frac{(n+2)(n+1)n}{3\times 2} [xn]F(x)=(3n+2)=3×2(n+2)(n+1)n

Devu and Flowers

对于每一朵花：
Fi(x)=1+x+x2+...+xfi=11−x−xfi+11−x=1−xfi+11−xF_i(x)=1+x+x^2+...+x^{f_i}=\frac{1}{1-x}-\frac{x^{f_i+1}}{1-x}=\frac{1-x^{f_i+1}}{1-x} Fi(x)=1+x+x2+...+xfi=1−x1−1−xxfi+1=1−x1−xfi+1
总的：
F(x)=F1(x)F2(x)...Fn(x)=∏i=1n(1−xfi+1)(1−x)n=A(x)⋅1(1−x)nF(x)=F_1(x)F_2(x)...F_n(x)=\frac{\prod_{i=1}^n(1-x^{f_i+1})}{(1-x)^n}=A(x)\cdot \frac{1}{(1-x)^n} F(x)=F1(x)F2(x)...Fn(x)=(1−x)n∏i=1n(1−xfi+1)=A(x)⋅(1−x)n1
A(x)A(x)A(x)可以通过dfs求解，O(2n)O(2^n)O(2n)。
故：
[xs]F(x)=∑i≥0[xi]A(x)⋅[xs−i]1(1−x)n=∑i≥0[xi]A(x)⋅(s−i+n−1n−1)\begin{matrix} [x^s]F(x)&=&\sum_{i\ge 0}[x^i]A(x)\cdot [x^{s-i}]\frac{1}{(1-x)^n} \\ \\ &=& \sum_{i\ge 0}[x^i]A(x)\cdot \binom{s-i+n-1}{n-1} \end{matrix} [xs]F(x)==∑i≥0[xi]A(x)⋅[xs−i](1−x)n1∑i≥0[xi]A(x)⋅(n−1s−i+n−1)

Sweets

对于每一个糖果：
Fi(x)=1+x+x2+...+xmi=11−x−xmi+11−x=1−xmi+11−xF_i(x)=1+x+x^2+...+x^{m_i}=\frac{1}{1-x}-\frac{x^{m_i+1}}{1-x}=\frac{1-x^{m_i+1}}{1-x} Fi(x)=1+x+x2+...+xmi=1−x1−1−xxmi+1=1−x1−xmi+1
总的：
F(x)=F1(x)F2(x)...Fn(x)=∏i=1n(1−xmi+1)(1−x)n=A(x)⋅1(1−x)nF(x)=F_1(x)F_2(x)...F_n(x)=\frac{\prod_{i=1}^n(1-x^{m_i+1})}{(1-x)^n}=A(x)\cdot \frac{1}{(1-x)^n} F(x)=F1(x)F2(x)...Fn(x)=(1−x)n∏i=1n(1−xmi+1)=A(x)⋅(1−x)n1

∑s=ab[xs]F(x)=∑s=ab∑i≥0[xi]A(x)⋅(s−i+n−1n−1)=∑i≥0[xi]A(x)⋅∑s=ab(s−i+n−1n−1)\begin{matrix} \sum_{s=a}^{b}[x^s]F(x)&=&\sum_{s=a}^{b} \sum_{i\ge 0}[x^i]A(x)\cdot \binom{s-i+n-1}{n-1} \\ \\ &=& \sum_{i\ge 0}[x^i]A(x)\cdot \sum_{s=a}^{b} \binom{s-i+n-1}{n-1} \end{matrix} ∑s=ab[xs]F(x)==∑s=ab∑i≥0[xi]A(x)⋅(n−1s−i+n−1)∑i≥0[xi]A(x)⋅∑s=ab(n−1s−i+n−1)

因为：
(aj)+(aj+1)=(a+1j+1)(a+1j)+(a+1j+1)=(a+2j+1)(a+2j)+(a+2j+1)=(a+3j+1)...(bj)+(bj+1)=(b+1j+1)\begin{matrix} \binom{a}{j}+\binom{a}{j+1}&=&\binom{a+1}{j+1} \\ \\ \binom{a+1}{j}+\binom{a+1}{j+1}&=&\binom{a+2}{j+1} \\ \\ \binom{a+2}{j}+\binom{a+2}{j+1}&=&\binom{a+3}{j+1} \\ \\ ... \\ \\ \binom{b}{j}+\binom{b}{j+1}&=&\binom{b+1}{j+1} \\ \\ \end{matrix} (ja)+(j+1a)(ja+1)+(j+1a+1)(ja+2)+(j+1a+2)...(jb)+(j+1b)====(j+1a+1)(j+1a+2)(j+1a+3)(j+1b+1)
(aj)+(a+1j)+(a+2j)+...+(bj)=(b+1j+1)−(aj+1)\binom{a}{j}+\binom{a+1}{j}+\binom{a+2}{j}+...+\binom{b}{j}=\binom{b+1}{j+1}-\binom{a}{j+1} (ja)+(ja+1)+(ja+2)+...+(jb)=(j+1b+1)−(j+1a)

故：
∑s=ab[xs]F(x)=∑i≥0[xi]A(x)⋅((b−i+n−1+1n−1+1)−(a−i+n−1n−1+1))\sum_{s=a}^{b}[x^s]F(x)=\sum_{i\ge 0}[x^i]A(x)\cdot (\binom{b-i+n-1+1}{n-1+1}-\binom{a-i+n-1}{n-1+1}) s=a∑b[xs]F(x)=i≥0∑[xi]A(x)⋅((n−1+1b−i+n−1+1)−(n−1+1a−i+n−1))

Blocks

黄色和绿色只能有偶数个：
F(x)=1+x22!+x44!+...=exp⁡(x)+exp⁡(−x)2F(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+...=\frac{\exp(x)+\exp(-x)}{2} F(x)=1+2!x2+4!x4+...=2exp(x)+exp(−x)
蓝色和黄色可以有任意个：
G(x)=1+x+x22!+x33!+...=exp⁡(x)G(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...=\exp(x) G(x)=1+x+2!x2+3!x3+...=exp(x)

总的：
H(x)=F(x)2G(x)2=(exp⁡(x)+exp⁡(−x)2)2exp⁡(x)2=exp⁡(x)2+2+exp⁡(−x)24exp⁡(x)2=exp⁡(x)4+2exp⁡(x)2+14=exp⁡(4x)+2exp⁡(2x)+14=∑n>04n+2⋅2n4xnn!+14\begin{matrix} H(x)&=&F(x)^2G(x)^2 \\ \\ &=& (\frac{\exp(x)+\exp(-x)}{2})^2\exp(x)^2 \\ \\ &=& \frac{\exp(x)^2+2+\exp(-x)^2}{4}\exp(x)^2 \\ \\ &=& \frac{\exp(x)^4+2\exp(x)^2+1}{4} \\ \\ &=& \frac{\exp(4x)+2\exp(2x)+1}{4} \\ \\ &=& \sum_{n>0}\frac{4^n+2\cdot 2^n}{4}\frac{x^n}{n!}+\frac{1}{4} \end{matrix} H(x)======F(x)2G(x)2(2exp(x)+exp(−x))2exp(x)24exp(x)2+2+exp(−x)2exp(x)24exp(x)4+2exp(x)2+14exp(4x)+2exp(2x)+1∑n>044n+2⋅2nn!xn+41
故，结果为：
n!⋅[xn]H(x)=14(4n+2⋅2n)=4n−1+2n−1n!\cdot [x^n]H(x)=\frac{1}{4}(4^n+2\cdot 2^n)=4^{n-1}+2^{n-1} n!⋅[xn]H(x)=41(4n+2⋅2n)=4n−1+2n−1

Lust

对于每一次操作：
res+=(a1a2...ax...an)−a1a2...(ax−1)...anres += (a_1a_2 ... a_x ... a_n)-a_1 a_2 ... (a_x-1) ... a_n res+=(a1a2...ax...an)−a1a2...(ax−1)...an
故，最终resresres会变为：(其中xix_ixi为aia_iai被减的次数）
res+=(a1a2...ax...an)−(a1−x1)(a2−x2)...(an−xn)res += (a_1a_2 ... a_x ... a_n)-(a_1-x_1)(a_2-x_2)...(a_n-x_n) res+=(a1a2...ax...an)−(a1−x1)(a2−x2)...(an−xn)

E[(a1−x1)(a2−x2)...(an−xn)]=1nk∑x1+x2+...+xn=k(a1−x1)(a2−x2)...(an−xn)(kx1,x2,..xn)=1nk∑x1+x2+...+xn=k(a1−x1)(a2−x2)...(an−xn)k!x1!x2!..xn!=k!nk∑x1+x2+...+xn=ka1−x1x1!a2−x2x2!...an−xnxn!\begin{matrix} &&\mathbb{E} [(a_1-x_1)(a_2-x_2)...(a_n-x_n)] \\ \\ &=&\frac{1}{n^k} \sum_{x_1+x_2+...+x_n=k}(a_1-x_1)(a_2-x_2)...(a_n-x_n)\binom{k}{x_1,x_2,..x_n} \\ \\ &=&\frac{1}{n^k} \sum_{x_1+x_2+...+x_n=k}(a_1-x_1)(a_2-x_2)...(a_n-x_n)\frac{k!}{x_1!x_2!..x_n!} \\ \\ &=&\frac{k!}{n^k}\sum_{x_1+x_2+...+x_n=k}\frac{a_1-x_1}{x_1!}\frac{a_2-x_2}{x_2!}...\frac{a_n-x_n}{x_n!} \\ \\ \end{matrix} ===E[(a1−x1)(a2−x2)...(an−xn)]nk1∑x1+x2+...+xn=k(a1−x1)(a2−x2)...(an−xn)(x1,x2,..xnk)nk1∑x1+x2+...+xn=k(a1−x1)(a2−x2)...(an−xn)x1!x2!..xn!k!nkk!∑x1+x2+...+xn=kx1!a1−x1x2!a2−x2...xn!an−xn

令:
Fi(x)=∑j≥0ai−jj!xj=∑j≥0aij!xj−∑j≥0jj!xj=aiexp⁡(x)−xexp⁡(x)F_i(x)=\sum_{j\ge 0}\frac{a_i-j}{j!}x^j=\sum_{j\ge 0}\frac{a_i}{j!}x^j-\sum_{j\ge 0}\frac{j}{j!}x^j= a_i\exp(x)-x\exp(x) Fi(x)=j≥0∑j!ai−jxj=j≥0∑j!aixj−j≥0∑j!jxj=aiexp(x)−xexp(x)
故
F(x)=F1(x)F2(x)...Fn(x)=∏i=1n(ai−x)exp⁡(x)n=∏i=1n(ai−x)exp⁡(nx)=A(x)exp⁡(nx)F(x)=F_1(x)F_2(x)...F_n(x)=\prod_{i=1}^n(a_i-x)\exp(x)^n=\prod_{i=1}^n(a_i-x)\exp(nx)=A(x)\exp(nx) F(x)=F1(x)F2(x)...Fn(x)=i=1∏n(ai−x)exp(x)n=i=1∏n(ai−x)exp(nx)=A(x)exp(nx)

[xk]F(x)=∑i=0k[xi]A(x)⋅[xk−i]exp⁡(nx)=∑i=0n[xi]A(x)⋅[xk−i]exp⁡(nx)=∑i=0n[xi]A(x)⋅nk−i(k−i)!\begin{matrix} [x^k]F(x)&=&\sum_{i=0}^k[x^i]A(x)\cdot [x^{k-i}]\exp(nx)\\ \\ &=&\sum_{i=0}^n[x^i]A(x)\cdot [x^{k-i}]\exp(nx) \\ \\ &=&\sum_{i=0}^n[x^i]A(x)\cdot\frac{n^{k-i}}{(k-i)!} \end{matrix} [xk]F(x)===∑i=0k[xi]A(x)⋅[xk−i]exp(nx)∑i=0n[xi]A(x)⋅[xk−i]exp(nx)∑i=0n[xi]A(x)⋅(k−i)!nk−i
结果为：
E(res)=∑i=1nai−k!nk∑i=0n[xk]F(x)=∑i=1nai−∑i=0n[xi]A(x)⋅ki‾ni\mathbb{E} (res) = \sum_{i=1}^na_i-\frac{k!}{n^k}\sum_{i=0}^n[x^k]F(x)=\sum_{i=1}^na_i-\sum_{i=0}^n[x^i]A(x)\cdot \frac{k^{\underline{i}}}{n^i} E(res)=i=1∑nai−nkk!i=0∑n[xk]F(x)=i=1∑nai−i=0∑n[xi]A(x)⋅niki