Value of the bzoj5093 chart "the second class of Stirling number +ntt"

Problem Solving Ideas:

Consider directly enumerating the edges of each point, and the other points in the graph are connected to each other directly, then push the formula straight: Ans=n∗2c2n−1∑i=0n−1cin−1ik ans=n*2^{c_{n-1}^2}\sum\limits_{i=0}^{n-1}c_{n-1 }^ii^k

Use n n to represent n−1 n-1, while IK i^k can be represented by the second class of Stirling numbers: ans= (n+1) ∗2c2n∑i=0ncin∑j=0isjkcjij! ans= (n+1) *2^{c_{n}^2}\sum\limits_{i= 0}^{n}c_{n}^i\sum\limits_{j=0}^is_k^jc_i^jj!

Exchange I,j I,j order, with: ans= (n+1) ∗2c2n∑j=0nsjkj!∑i=jncincji