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\(\begin{align*} \sum\limits_{i=1}^n \binom{n}{i} i^k & = \sum\limits_{i=1}^n \binom{n}{i} \sum\limits_{j=1}^k \binom{i}{j} \left\{\begin{array}{cccc} k \\ j \end{array}\right\}j! \\ &= \sum\limits_{j=1}^k \left\{\begin{array}{cccc} k \\ j \end{array} \right\}j! \sum\limits_{i=1}^n \binom{n}{i} \binom{i}{j} \\ &= \sum\limits_{j=1}^k \left\{\begin{array}{cccc} k \\ j \end{array} \right\}j! \binom{n}{j} \sum\limits_{i=1}^n \binom{n-j}{i-j} \\ &= \sum\limits_{j=1}^k \left\{\begin{array}{cccc} k \\ j \end{array} \right\}j! \binom{n}{j} \sum\limits_{i=0}^{n-j} \binom{n-j}{i} \\ &= \sum\limits_{j=1}^k \left\{\begin{array}{cccc} k \\ j \end{array} \right\}j! \binom{n}{j} 2^{n-j} \end{align*}\)

预处理斯特林数,边枚举\(j\)边维护组合数和阶乘。

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转载自www.cnblogs.com/Itst/p/10842919.html
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