题意：(k) 种球，每种个数必须是 (d) 的倍数，共 (n) 个，求排成一行的方案数
[egin{align*}
ext{Ans}&=n!sum_{i_1 ge 0,dmid i_1}frac{1}{i_1!}sum_{0le i_2,dmid i_2}frac{1}{i_2!}cdotssum_{0le i_k,dmid i_k}frac{1}{i_k!}\
&=[x^n]n!(sum_{ige 0,dmid i}frac{x^i}{i!})^k\
end{align*}
]
[egin{align*}
sum_{ige 0,dmid i}frac{x^i}{i!}&=sum_{ige 0}frac{x^i}{i!}frac{1}{d}sum_{0le j < d}{omega_d^j}^i& ext{单位根反演}\
&=frac{1}{d}sum_{0le j < d}sum_{ige 0}frac{(xomega_d^j)^i}{i!}\
&=frac{1}{d}sum_{0le j < d}e^{xomega_d^j}& ext{泰勒展开}\
end{align*}
]
(d=2)：
[egin{align*}
sum_{ige 0,dmid i}frac{x^i}{i!}&=frac{1}{d}sum_{0le j < d}e^{xomega_d^j}\
&=frac{e^x+e^{-x}}{2}\
(sum_{ige 0,dmid i}frac{x^i}{i!})^k&=n!(frac{e^x+e^{-x}}{2})^k\
&=frac{1}{2^k}sum_{0le ile k}C_k^ie^{ix}e^{-x(k-i)} & ext{二项式定理}\
&=frac{1}{2^k}sum_{0le ile k}C_k^ie^{2ix-kx}\
ext{Ans}&=[x^n]n!(sum_{ige 0,dmid i}frac{x^i}{i!})^k\
&=frac{n!}{2^k}sum_{0le ile k}C_k^i[x^n]e^{2ix-kx}\
&=frac{n!}{2^k}sum_{0le ile k}C_k^ifrac{(2i-k)^n}{n!} & ext{泰勒展开}\
&=frac{1}{2^k}sum_{0le ile k}C_k^i(2i-k)^n
end{align*}
]
(d=3)：
[egin{align*}
sum_{ige 0,dmid i}frac{x^i}{i!}&=frac{1}{d}sum_{0le j < d}e^{xomega_d^j}\
&=frac{e^x+e^{omega_3x}+e^{omega_3^2x}}{3}\
(sum_{ige 0,dmid i}frac{x^i}{i!})^k&=n!(frac{e^x+e^{omega_3x}+e^{omega_3^2x}}{3})^k\
&=frac{1}{3^k}sum_{0le ile k}C_k^isum_{0le jle k-i}C_{k-i}^je^{ix}e^{jomega_3x}e^{(k-i-j)omega_3^2x} & ext{多项式定理}\
&=frac{1}{3^k}sum_{0le ile k}sum_{0le jle k-i}C_k^iC_{k-i}^je^{ix+jomega_3x+komega_3^2x-iomega_3^2x-jomega_3^2x}\
ext{Ans}&=[x^n]n!(sum_{ige 0,dmid i}frac{x^i}{i!})^k\
&=frac{n!}{3^k}sum_{0le ile k}sum_{0le jle k-i}C_k^iC_{k-i}^j[x^n]e^{ix+jomega_3x+komega_3^2x-iomega_3^2x-jomega_3^2x}\
&=frac{n!}{3^k}sum_{0le ile k}sum_{0le jle k-i}C_k^iC_{k-i}^jfrac{(i+jomega_3+komega_3^2-iomega_3^2-jomega_3^2)^n}{n!} & ext{泰勒展开}\
&=frac{1}{3^k}sum_{0le ile k}sum_{0le jle k-i}C_k^iC_{k-i}^j(i+jomega_3+komega_3^2-iomega_3^2-jomega_3^2)^n
end{align*}
]