【分析】
单位根反演 + CZT
\(\begin{aligned} \sum_{i=0}^n\binom n i p^i\lfloor{i\over k}\rfloor&=\sum_{i=0}^n\binom n i p^i\cdot {i-(i\bmod k)\over k} \\&={1\over k}\left(\ p\sum_{i=0}^n \binom n i {\text d\over \text dp}p^i - \sum_{r=0}^{k-1}\sum_{i=0}^n \binom n i p^i r [k\mid (i-r)]\ \right) \\&={1\over k}\left(\ p\cdot {\text d\over \text dp}(p+1)^n-\sum_{r=0}^{k-1}\sum_{i=0}^n \binom n i p^i\cdot {1\over k}\sum_{t=0}^{k-1}\omega_k^{(i-r)t}\ \right)&\text{(二项式定理、单位根反演)} \end{aligned}\)