[CSP-S2019 江西] 和积和

$$S(l,r)=\sum _{i=l}^r{a}_i\times \sum _{i=l}^r{b}_i$$$$\mathrm{求}\sum _{l=1}^n\sum _{r=l}^{n}S(l,r)$$$$\mathrm{求}\sum _{l=1}^n\sum _{r=l}^n(\sum _{i=l}^r{a}_i\times \sum _{i=l}^r{b}_i)$$$$\begin{gathered} S{A}_i=\sum _{j=1}^i{a}_j \\ S{B}_i=\sum _{j=1}^i{b}_j \end{gathered}$$$$\mathrm{求}\sum _{l=1}^n\sum _{r=l}^n(S{A}_r-S{A}_{l-1})\times (S{B}_r-S{B}_{l-1})$$$$\begin{gathered} \mathrm{求} \\ \sum _{l=1}^n\sum _{r=l}^n(S{A}_r\ast S{B}_r) \\ -\sum _{l=1}^n\sum _{r=l}^n(S{A}_r\ast S{B}_{l-1}) \\ -\sum _{l=1}^n\sum _{r=l}^n(S{A}_{l-1}\ast S{B}_r) \\ +\sum _{l=1}^n\sum _{r=l}^n(S{A}_{l-1}\times S{B}_{l-1}) \end{gathered}$$$$\begin{gathered} SS{A}_i=\sum _{j=1}^{i}S{A}_j \\ SS{B}_i=\sum _{j=1}^{i}S{B}_j \\ SSA{B}_i=\sum _{j=1}^{i}S{A}_j\times S{B}_j \end{gathered}$$$$\begin{gathered} \mathrm{求} \\ \sum _{l=1}^n(SSA{B}_n-SSA{B}_{l-1}) \\ -\sum _{l=1}^n(SS{A}_n-SS{A}_{l-1})\times S{B}_{l-1} \\ -\sum _{l=1}^{n}S{A}_{l-1}\times (SS{B}_n-SS{B}_{l-1}) \\ +\sum _{l=1}^n(n-l+1)\times (S{A}_{l-1}\times S{B}_{l-1}) \end{gathered}$$