直接帶入
\[\begin{aligned}
\sum_{i=0}^{d}b_ix^i&=\sum_{i=0}^{d}a_i(x+s)^{i}\\
&=\sum_{i=0}^{d}x_i\sum_{j=i}^{d}\binom{j}{i}a_js^{j-i}\\
\end{aligned}
\]
\[b_i=\sum_{j=i}^{d}\binom{j}{i}a_js^{j-i}
\]
觀察一些特殊值,發現:
\[s=\frac{b_{d-1}-a_{d-1}\times d}{a_d}
\]
觀察拉格朗日插值的式子,發現
\[[x^d]\sum_{i=0}^{d}y_i\prod_{j\ne i}\frac{x-x_j}{x_i-x_j}=\sum_{i=0}^{d}\frac{y_i}{\prod_{j\ne i}(x_i-x_j)}\\
[x^{d-1 }]\sum_{i=0}^{d}y_i\prod_{j\ne i}\frac{x-x_j}{x_i-x_j}=\sum_{i=0}^{d}(\frac{y_i}{\prod_{j\ne i}(x_i-x_j)}\times \sum_{j\ne i} x_j)
\]
這題橫座標是 \(0\sim d\) 所以可以直接 \(O(d)\) 預處理階乘和階乘逆元。