求：
\(S(n,m)=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}lcm(i,j)\)
显然：
\(S(n,m)=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}\frac{ij}{gcd(i,j)}\)
枚举g：
\(S(n,m)=\sum\limits_{g=1}^{n}\frac{1}{g}\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}ij[gcd(i,j)==g]\)
除以g：
\(S(n,m)=\sum\limits_{g=1}^{n}g\sum\limits_{i=1}^{\lfloor\frac{n}{g}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{g}\rfloor}ij[gcd(i,j)==1]\)
记：
\(S_1(n,m)=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}ij[gcd(i,j)==1]\)
原式：
\(S(n,m)=\sum\limits_{g=1}^{n}gS_1(\lfloor\frac{n}{g}\rfloor,\lfloor\frac{m}{g}\rfloor)\)
化简\(S_1(n,m)\)，显然：
\(S_1(n,m)=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}ij\sum\limits_{k|gcd(i,j)}\mu(k)\)
枚举k：
\(S_1(n,m)=\sum\limits_{k=1}^{min}\mu(k)\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}ij[k|gcd(i,j)]\)
显然：
\(S_1(n,m)=\sum\limits_{k=1}^{min}\mu(k)\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}ij[k|i][k|j]\)
This time can be divided by K:
\ (S_1 (n-, m) = \ SUM \ limits_. 1} ^ {K = min {} \ MU (K) 2 K ^ \ SUM \ limits_. 1 = {I} ^ {\ lfloor \ frac {n} {k } \ rfloor} \ sum \ limits_ {j = 1} ^ {\ lfloor \ frac {m} {k} \ rfloor} ij [1 | i] [1 | j] \)
即：
\(S_1(n,m)=\sum\limits_{k=1}^{min}\mu(k)k^2\sum\limits_{i=1}^{\lfloor\frac{n}{k}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{k}\rfloor}ij\)
记：
\(S_2(n,m)=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}ij\)
原式：
\(S_1(n,m)=\sum\limits_{k=1}^{min}\mu(k)k^2S_2(\lfloor\frac{n}{k}\rfloor,\lfloor\frac{m}{k}\rfloor)\)
显然：
\(S_2(n,m)=\sum\limits_{i=1}^{n}i\sum\limits_{j=1}^{m}j\)
即：
\(S_2(n,m)=\frac{1}{4}n(n+1)m(m+1)\)
Time complexity:
seeking \ (S_2 (n, m) \) is \ (O (. 1) \) , block request \ (S_1 (n, m) \) is \ (O (n ^ {\ frac {1 {2}}}) \) (presumably), find block \ (S (n, m) \) is \ (O (n-) \) (approximate).
Linear need screening out: \ (\ SUM \ limits_. 1} ^ {K = min {} \ MU (K) 2 K ^ \)