【题解】「2018 集训队互测 Day 3」蒜头的奖杯

题目链接

题目大意：给定\(n,A,B,C,D,E,F\)，求

\[\sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^nA_iB_jC_kD_{\gcd(i,j)}E_{\gcd(i,k)}F_{\gcd(j,k)} \]

首先容易想到三个同时算的方法：对\(D,E,F\)进行反演，得

\[\sum_{i=1}^nA_i\sum_{p|i}D_p\sum_{p|j}^nB_j\sum_{r|j}F_r\sum_{r|k}^nC_k\sum_{q|k}E_q[i\%q=0] \]

设数组\(tmp\)初始值为\(A\)，可以先变换再与一个函数对应位置相乘，进行\(5\)次。要分别对\(1\text{~}n\)的每一个起始位置算一次，\(O(n^2\log n)\)，喜提TLE。

三个同时算不好算，考虑拆开算，原式即

\[\sum_{i=1}^n\sum_{j=1}^nA_iB_jD_{\gcd(i,j)}\sum_{k=1}^nC_kE_{\gcd(i,k)}F_{\gcd(j,k)} \]

同时对\(E\)和\(F\)进行莫比乌斯反演得

\[\sum_{i=1}^n\sum_{j=1}^nA_iB_jD_{\gcd(i,j)}\sum_{k=1}^nC_k\sum_{q|i,q|k}E_q\sum_{r|j,r|k}F_r \]

于是，根据套路，将被拆开的\(C,E,F\)提到前面，再提出下标的\(\gcd\)

\[\begin{align} &\;\;\;\;\sum_{i=1}^n\sum_{j=1}^nA_iB_jD_{\gcd(i,j)}\sum_{k=1}^nC_k\sum_{q|i,q|k}E_q\sum_{r|j,r|k}F_r\\ &=\sum_{k=1}^nC_k\sum_{q|k}E_q\sum_{r|k}F_r\sum_{i=1}^{\lfloor\frac nq\rfloor}\sum_{j=1}^{\lfloor\frac nr\rfloor}A_{iq}B_{jr}D_{\gcd(iq,jr)}\\ &=\sum_{s=1}^n\sum_{q=1}^{\lfloor\frac ns\rfloor}\sum_{r=1}^{\lfloor\frac n{sq}\rfloor}[\gcd(q,r)=1]E_{sq}F_{sr}\sum_{k=1}^{\lfloor\frac n{sqr}\rfloor}C_{sqrk}\sum_{i=1}^{\lfloor\frac n{sq}\rfloor}\sum_{j=1}^{\lfloor\frac n{sr}\rfloor}A_{isq}B_{jsr}D_{s\cdot\gcd(iq,jr)} \end{align}\]

设\(C'_i=\sum_{j=1}^{\lfloor\frac ni\rfloor}C_{ij},A'_i=A_{si},B'_i=B_{si},D'_i=D_{si},m=\lfloor\frac ns\rfloor\)，则：

\[\begin{align} &\;\;\;\;\sum_{s=1}^n\sum_{q=1}^{\lfloor\frac ns\rfloor}\sum_{r=1}^{\lfloor\frac n{sq}\rfloor}[\gcd(q,r)=1]E_{sq}F_{sr}\sum_{k=1}^{\lfloor\frac n{sqr}\rfloor}C_{sqrk}\sum_{i=1}^{\lfloor\frac n{sq}\rfloor}\sum_{j=1}^{\lfloor\frac n{sr}\rfloor}A_{isq}B_{jsr}D_{s\cdot\gcd(iq,jr)}\\ &=\sum_{s=1}^n\sum_{q=1}^{\lfloor\frac ns\rfloor}\sum_{r=1}^{\lfloor\frac n{sq}\rfloor}[\gcd(q,r)=1]E_{sq}F_{sr}C'_{sqr}\sum_{i=1}^{\lfloor\frac mq\rfloor}\sum_{j=1}^{\lfloor\frac mr\rfloor}A'_{iq}B'_{jr}D'_{\gcd(iq,jr)} \end{align}\]

将\(D'\)反演，得

\[\begin{align} &\;\;\;\;\sum_{s=1}^n\sum_{q=1}^{\lfloor\frac ns\rfloor}\sum_{r=1}^{\lfloor\frac n{sq}\rfloor}[\gcd(q,r)=1]E_{sq}F_{sr}C'_{sqr}\sum_{i=1}^{\lfloor\frac mq\rfloor}\sum_{j=1}^{\lfloor\frac mr\rfloor}A'_{iq}B'_{jr}\sum_{p|iq,p|jr}D'_p\\ &=\sum_{s=1}^n\sum_{q=1}^{\lfloor\frac ns\rfloor}\sum_{r=1}^{\lfloor\frac n{sq}\rfloor}[\gcd(q,r)=1]E_{sq}F_{sr}C'_{sqr}\sum_{q|i}^mA'_i\sum_{p|i}D'_p\sum_{p|j}B'_j[j\%r=0] \end{align}\]

借鉴之前的思路，枚举\(s,q\)，设\(tmp_i=A'_i[i\%q=0]\)，对\(tmp\)进行变换\(tmp'_i=D'_i\sum_{d|i}tmp_i\)得\(tmp'\)，再对\(tmp'\)进行变换\(tmp''_d=B'_d\sum_{d|i}tmp'_i\)得\(tmp''\)。

再枚举\(r\)，则\(s,q,r\)对答案的贡献为\(E_{sq}F_{sr}C'_{sqr}\sum_{i=1}^{\lfloor\frac mr\rfloor}tmp''_{ir}\)

注意常数优化

代码太丑不贴了……