Educational Codeforces Round 78 (Rated for Div. 2) F. Cards 第二类斯特林数
题意
给你洗 n n n次牌，这副牌有m张牌，其中有一个大王，你每次会抽出最顶的一张然后看完之后就放回去，假设你看过x次大王，你的贡献就是x^k，求期望贡献
分析
题目就是要求
= ∑ i = 1 n ( n i ) i k ( m − 1 ) n − i m n =\frac{\sum\limits_{i=1}^{n} \binom{n}{i} i^k(m-1)^{n-i}}{m^n} =mni=1∑n(in)ik(m−1)n−i
然后一看这种形式大多就用第二斯特林数展开
n k = ∑ i = 0 k S ( k , i ) i ! ( n i ) n^k = \sum\limits_{i=0}^{k}S(k,i)i!\binom{n}{i} nk=i=0∑kS(k,i)i!(in)
= ∑ i = 1 n ( n i ) ∑ j = 0 m i n ( i , k ) S ( k , j ) j ! ( i j ) ( m − 1 ) n − i m n =\frac{\sum\limits_{i=1}^{n}\binom{n}{i}\sum\limits_{j=0}^{min(i,k)}S(k,j)j!\binom{i}{j}(m-1)^{n-i}}{m^n} =mni=1∑n(in)j=0∑min(i,k)S(k,j)j!(ji)(m−1)n−i
= ∑ j = 0 m i n ( n , k ) S ( k , j ) ∑ i = j n n ! ( n − i ) ! ( i − j ) ! ( m − 1 ) n − i m n =\frac{\sum\limits_{j=0}^{min(n,k)}S(k,j)\sum\limits_{i=j}^{n} \frac{n!}{(n-i)!(i-j)!} (m-1)^{n-i}}{m^n} =mnj=0∑min(n,k)S(k,j)i=j∑n(n−i)!(i−j)!n!(m−1)n−i
考虑n很大，把后面带n的部分去掉
= ∑ j = 0 m i n ( n , k ) S ( k , j ) n ! ( n − j ) ! ∑ i = 0 n − j ( n − j i ) ( m − 1 ) n − i − j m n =\frac{\sum\limits_{j=0}^{min(n,k)}S(k,j)\frac{n!}{(n-j)!}\sum\limits_{i=0}^{n-j}\binom{n-j}{i}(m-1)^{n-i-j}}{m^n} =mnj=0∑min(n,k)S(k,j)(n−j)!n!i=0∑n−j(in−j)(m−1)n−i−j
用二项式定理合并之后就行了
= ∑ j = 0 m i n ( n , k ) S ( k , j ) n ! ( n − j ) ! m n − j m n =\frac{\sum\limits_{j=0}^{min(n,k)}S(k,j)\frac{n!}{(n-j)!}m^{n-j}}{m^n} =mnj=0∑min(n,k)S(k,j)(n−j)!n!mn−j
代码
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