数学

由引理$ans= \prod_{j=1}^{ \lceil \frac{m}{2} \rceil} \prod_{k=1}^{ \lceil \frac{n}{2} \rceil} 4( \cos^2 \frac{ \pi j }{m+1} + \cos^2 \frac{\pi k }{n+1} )$wiki
可知

$$ans= \prod_{j=1}^{ \lceil \frac{m}{2} \rceil} \prod_{k=1}^{ \lceil \frac{n}{2} \rceil} 4( \cos^2 \frac{ \pi j }{m+1} + \cos^2 \frac{\pi k }{n+1} ) \\ =[\prod_{j=1}^{m} \prod_{k=1}^{n} 4( \cos^2 \frac{ \pi j }{m+1} + \cos^2 \frac{\pi k }{n+1} )]^{\frac {1} {4} } \\ =[\prod_{j=1}^{m} \prod_{k=1}^{n} \left| 2(i \cos\frac {\pi j} {m+1}-\cos\frac {\pi k} {n+1} )\right| ^2 ]^{\frac {1}{4}} \\ $$

引入第二类Chebyshev多项式 mathworld

$$U_n(x)= 2^n \prod_{k=1}^{n} (x- \cos\frac{\pi k} {n+1}) \\ 于是有ans= \sqrt{ \left | \prod_{j=1}^m U_n(i \cos\frac {\pi j} {m+1}) \right | } \\ 记n阶矩阵B_n为除了主对角线边上元素为1，其余为0的矩阵 \\ 也就是Bn_{ij}在\left | i-j \right |=1时为1，其余时候为0 \\ 第二类Chebyshev多项式可写作\det(2x \times E_n + B_n) \\ 其中E_n为n阶单位矩阵 \\ 于是ans= \sqrt{ \left | \det[\prod_{j=1}^m (2i \cos\frac {\pi j} {m+1} I_n+B_n)] \right | } \\ \\ =\sqrt{ \left | \det[2^m \prod_{j=1}^m (i\frac {B_n}{2}-\cos\frac{\pi j} {m+1}I_n)] \right | } \\ \\ =\sqrt{ \left | \det[U_m(i \frac{B_n}{2})] \right | }$$

$$由Clenshaw递推公式,第二类Chebyshev多项式满足递推式 \\ \left\{\begin{array}{l} U_0(x)=1 \\ U_1(x)=2x; \\ U_{n+1}(x)=2xU_n(x)-U_{n-1}(x) \end{array}\right.$$

于是可求答案。