王总不屑于发，作为王总的小迷弟，我通过一些渠道拿到了王总的手迹并准备发出这珍贵的板书

$\vec{v}_1,\vec{v}_2,\cdots,\vec{v}_n$

$$\vec{a},\vec{b}\rightarrow 2\vec{a}-\vec{b}$$

$$(-1)^{c_1}2^{d_1}\vec{v}_1+(-1)^{c_2}2^{d_2}\vec{v}_2+\cdots$$

$$f_{n,a,b}$$

$$f_{0,0,0}=1,ans=f_{n,1,0}$$

$$f_{n,a',b'}=\sum_{a,b}\dbinom{n}{a'-(a-b),b'+(a-b)}f_{n-a'-b',a,b}$$

$$\hat{f}_{,n,a,b}=\dfrac{f_{n,a,b}}{n!}$$

$$\hat{f}_{n,a,b}=\sum_{a',b'}\dfrac{f_{n-a-b,a',b'}}{(a-(a'-b'))!(b+(a'-b')!)}$$

$$g_{n,\Delta}=\sum_{a-b=\Delta}\hat{f}_{n,a,b}$$

$$g_{0,\Delta}=[\Delta=0],ans=n!(g_{n-1,0}+g_{n-1,1}),g_{n,-\Delta}=g_{n,\Delta}$$

$$\hat{f}_{n,a,b}=\sum_{\Delta'}\dfrac{g_{n-a-b,\Delta'}}{(a-\Delta')!(b+\Delta')!}$$ $$g_{n,\Delta}=\sum_{a-b=\Delta,\Delta'\atop a,b\in\mathbf{N},\Delta'\in\mathbf{Z}}\dfrac{g_{n-a-b,\Delta'}}{(a-\Delta')!(b+\Delta')!}$$

$$x=a-\Delta',y=b+\Delta'$$

$$g_{n,\Delta}=\sum_{\Delta'\in\mathbf{Z}\atop x,y\in\mathbf{N}}\dfrac{g_{n-x-y,\Delta'}}{x!y!}[x+\Delta'\ge 0][y-\Delta'\ge 0][2\Delta'=\Delta-(x-y)]$$

如果忽略 $[x+\Delta'\ge 0][y-\Delta'\ge 0]$，则有 $$g_{n,\Delta}=\sum_{x}\dfrac{1}{x!}\sum_{y,\Delta'}\dfrac{g_{(n-x)-y,\Delta'}}{y!}[2\Delta'=(\Delta-x)+y]$$ $$=\sum_{y}\dfrac{1}{y!}\sum_{x,\Delta'}\dfrac{g_{(n-y)-x,\Delta'}}{x!}[2\Delta'=(\Delta+y)-x]$$

显然设 $h_{n,\Delta}=\displaystyle\sum_{y,\Delta'}\dfrac{g_{n-y,\Delta'}}{y!}[2\Delta'=\Delta+y]$，则

$$g_{n,\Delta}=\sum_{x}\dfrac{1}{x!}h_{n-x,\Delta-x}$$ $$h_{n,\Delta}=\sum_{\Delta'\atop y=2\Delta'-\Delta}\dfrac{g_{n-(2\Delta'-\Delta),\Delta'}}{(2\Delta'-\Delta)!}$$

均可 $\Theta(n^3)$ 递推

不忽略的话 - 若 $\Delta'>0,\forall x\in\mathbf{N},x+\Delta'\ge0$ - 若 $\Delta'<0,\forall y\in\mathbf{N},y-\Delta'\ge 0$ - 若 $\Delta=0$，都可以忽略

于是设 $$h_{n,\Delta}=\sum_{\Delta'\ge 1}\dfrac{g_{n-(2\Delta'-\Delta),\Delta'}}{(2\Delta'-\Delta)!}[\Delta'\ge\Delta]$$

则有

$$g_{n,\Delta}=\sum_x\dfrac{h_{n-x,\Delta-x}}{x!}+\sum_y\dfrac{h_{n-y,-\Delta-y}}{y!}+\sum_{x-y=\Delta}\dfrac{g_{n-x-y,0}}{x!y!}$$