【bzoj4709】[Jsoi2011]柠檬

$$f_j + a_i {s_j}^2 - 2a_is_is_j > f_k + a_i {s_k}^2 - 2a_is_k$$ $$f_j - f_k + a_i{s_j}^2 - a_i{s_k}^2 > 2a_is_i\times{(s_j - s_k)}$$ $$\frac{f_j - f_k + a_i{s_j}^2 - a_i{s_k}^2}{2\times{(s_j - s_k)}} > a_is_i$$