P4240 毒瘤之神的考验 [莫比乌斯反演]

神仙莫比乌斯反演。

有个结论。

(varphi(i imes j) = huge{left(frac{varphi(i) imes varphi(j) imes gcd(i,j)}{varphi(gcd(i,j))} ight)})

首先假设 (nleq m)

(sum_i sum_j varphi(i imes j) = sum_i sum_j {large left(frac{varphi(i) imes varphi(j) imes gcd(i,j)}{varphi(gcd(i,j))} ight)})

枚举(gcd)！

(sum_d sum_i sum_j frac{varphi(i) imes varphi(j) imes gcd(i,j)}{varphi(gcd(i,j))}[gcd(i,j) = d])

然后
(sum_d {left(frac{d}{varphi(d)} ight)} sum_i^{frac{n}{d}} sum_j^{frac{m}{d}}varphi(i imes d) varphi(j imes d)[gcd(i,j) = 1])

发现 (gcd) 可以用 (mu) 搞

(sum_{d|x} mu(d) = [x=1])

(sum_d {left(frac{d}{varphi(d)} ight)}sum_i^{frac{n}{d}} sum_j^{frac{m}{d}}varphi(i imes d) varphi(j imes d){left(sum_{k|gcd(i,j)} mu(k) ight)})

然后顺手枚举一波 (k)！

(sum_d {left(frac{d}{varphi(d)} ight)} sum_{k}^{frac{n}{d}}mu(k) sum_{i}^{frac{n}{kd}}sum_j^{frac{m}{kd}}varphi(i imes kd) imes varphi(j imes kd))

改变枚举顺序，令 (T=k imes d)

(sum_T sum_{k|T} mu(frac{T}{k}) frac{d}{varphi(d)} sum_i^{frac{n}{T}} sum_j^{frac{m}{T}} varphi(i imes T) imes varphi(j imes T))

然后我们搞搞 (F(T) = sum_T sum_{k|T} mu(frac{T}{k}))

((T, n, m) = sum_i^{frac{n}{T}} sum_j^{frac{m}{T}} varphi(i imes T) imes varphi(j imes T))

发现其实可以令 (G(x, y) = sum_i^x varphi(i imes y))

(G(x, y) = G(x - 1, y) + varphi(xy))

然后整个式子就变成了 (sum_T F(T) imes G(frac{n}{T},T) imes G(frac{m}{T},T))

(G) 可以直接暴力算，因为这个玩意是调和级数的。

但是发现这个答案数组不能搞得太大，所以需要一个根号分治，(leq B) 的部分暴力，(geq B+1) 的整除分块。