草稿纸

1	2	3
3	2	1

\(U_{0/1}\) 表示选 \(t_i = 0/1\) 的任务，\(V_{0/1}\) 表示时间选 \(t_i = 0/1\)

代价为 \(\sum ([U_0] + [U_1]) + \sum \infty ([\lnot U_0][\lnot V_0] + [\lnot U_1][\lnot V_1]) + \sum \infty [V_0][V_1] + \sum ([V_0] + [V_1])\)

\[\begin{aligned} \frac{1 - x ^ 2 y ^ 2}{1 - x - y - x ^ 2 y - x y ^ 2 - x ^ 2 y ^ 2} \\ \frac{1 - x}{1 - y - (1 + y)x} \\ [x ^ {a - i}]\frac{1 - x}{1 - y} \sum \left( \frac{y + 1}{y - 1} \right) ^ k x ^ k \\ \\ \sum 2 ^ {u + v} (-1) ^ {i + j - u - v} \dbinom{i + j}{i} \dbinom{a - i}{u} \dbinom{a - i + u - 1}{a - i} \dbinom{b - j}{v} \dbinom{b - j + v - 1}{b - j} \end{aligned}\]

\[\mathop{\sqrt{\frac{\underline{\quad\;} 1}{1 \underline{\quad} 1}}} ^ {\quad 1} \frac{-\ |\quad }{\cup\ | \sqrt{}}\]

\[\begin{aligned} f_i = \sum_{j = 0} ^ i (2 ^ j - 1) ^ {i - j} \dbinom{i}{j} \\ ans_i = \sum_{j = 0} ^ {i / 2} \dbinom{i}{2j} f_{i - 2j} \end{aligned}\]

\(\ker, \Im\)

\[\begin{align*} \begin{bmatrix} 1 & & & & 1 & & & \\ & 1 & & & & i & & \\ & & 1 & & & & -1 & \\ & & & 1 & & & & -i \\ 1 & & & & 1 & & & \\ & 1 & & & & i & & \\ & & 1 & & & & -1 & \\ & & & 1 & & & & -i \end{bmatrix} \begin{bmatrix} 1 & & 1 & & & & & \\ & 1 & & -1 & & & & \\ 1 & & -1 & & & & & \\ & 1 & & -1 & & & & \\ & & & & 1 & & 1 & \\ & & & & & 1 & & -1 \\ & & & & 1 & & -1 & \\ & & & & & 1 & & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 & & & & & & \\ 1 & -1 & & & & & & \\ & & 1 & 1 & & & & \\ & & 1 & -1 & & & & \\ & & & & 1 & 1 & & \\ & & & & 1 & -1 & & \\ & & & & & & 1 & 1 \\ & & & & & & 1 & -1 \end{bmatrix} \begin{bmatrix} x_0 \\ x_4 \\ x_2 \\ x_6 \\ x_1 \\ x_5 \\ x_3 \\ x_7 \end{bmatrix} \end{align*}\]

\[\begin{align*} &\sum_{i=1}^n i(n-i)\dbinom ni \\ =&n\sum_{i=1}^n i\dbinom ni-\sum_{i=1}^n i^2\dbinom ni \\ =&n^2\sum_{i=1}^n \dbinom{n-1}{i-1}-n\sum_{i=1}^n i\dbinom{n-1}{i-1} \\ =&n\sum_{i=1}^n (n-i)\dbinom{n-1}{n-i} \\ =&n(n-1)\sum_{i=1}^n \dbinom{n-2}{n-i-1} \\ =&n(n-1)\sum_{i=1}^n \dbinom{n-2}{i-1} \\ =&n(n-1)2^{n-2} \end{align*}\]

\[\begin{align*} &F=3uF+uG+1,\ \ G=uvF+3uvG \\ &F=\frac{uG+1}{1-3u},\ \ (1-3u)G=uv(uG+1)+3(1-3u)uvG \\ &G-3uG=u^2vG+uv+3uvG-9u^2vG \\ &(1-3u-3uv+8u^2v)G=uv,\ \ G=\frac{uv}{1-3u-3uv+8u^2v} \\ &G = \frac{uv}{1-3u}\times\frac{1}{1-v(3u-8u^2)/(1-3u)} \\ &F-3uvF=3uF-9u^2vF+u^2vF+1-3uv \\ &(1-3u-3uv+8u^2v)F=1-3uv,\ \ F=\frac{1-3uv}{1-3u-3uv+8u^2v} \\ &F+G = \frac{1-2uv}{1-3u}\times\frac{1}{1-v(3u-8u^2)/(1-3u)} \\ &[u^nv^k](F+G) = [u^n]\left(\frac{1}{1-3u}\left(\frac{3u-8u^2}{1-3u}\right)^k-\frac{2u}{1-3u}\left(\frac{3u-8u^2}{1-3u}\right)^{k-1}\right) \\ &[u^nv^k](F+G) = [u^{n-k}]\left(\frac{(3-8u)^k}{(1-3u)^{k+1}}-\frac{2(3-8u)^{k-1}}{(1-3u)^k}\right) \\ &[u^nv^k](F+G) = ([u^{n-k}]-2[u^{n-k-1}])\frac{(3-8u)^{k-1}}{(1-3u)^{k+1}} \\ \end{align*}\]