题意：
\(d(x)\)为
\(x\)的约数个数，求
\(\sum_{i=1}^n\sum{j=1}^md(ij)\) 由结论得
\[ans=\sum_{i=1}^n\sum_{j=1}^m\sum_{x|i}\sum_{y|j}[GCD(x,y)==1]\] 我们换个方向，考虑
\(x, y\)的贡献，则
\[ans=\sum_{x=1}^n\sum_{y=1}^m\lfloor\frac{n}{x}\rfloor\lfloor\frac{m}{y}\rfloor[GCD(x, y)==1]\] 令
\[f(d)=\sum_{x=1}^n\sum_{y=1}^m\lfloor\frac{n}{x}\rfloor\lfloor\frac{m}{y}\rfloor[GCD(x, y)==d]\]\[g(d)=\sum_{d|n}f(d)\]\[=\sum_{x=1}^n\sum_{y=1}^m\lfloor\frac{n}{x}\rfloor\lfloor\frac{m}{y}\rfloor[d|GCD(x, y)]\]\[=\sum_{x=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{y=1}^{\lfloor\frac{m}{d}\rfloor}\lfloor\frac{n}{xd}\rfloor\lfloor\frac{m}{yd}\rfloor\]\[=(\sum_{x=1}^{\lfloor\frac{n}{d}\rfloor}\lfloor\frac{n}{xd}\rfloor)(\sum_{y=1}^{\lfloor\frac{m}{d}\rfloor}\lfloor\frac{m}{yd}\rfloor)\]\[f(d)=\sum_{d|n}\mu(\frac{n}{d})g(n)\] 则
\(ans=f(1)=\sum_{d=1}^n\mu(d)g(d)\) 令
\(sum(n)=\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor\) 则
\(ans=\sum_{d=1}^n\mu(d)*sum(\lfloor\frac{n}{d}\rfloor)*sum(\lfloor\frac{m}{d}\rfloor)\) 不开
\(O_2\)会
\(T\)掉两个点
void init(){ miu[1]=1; for(int i=2; i < Maxn; i++){ if(!p[i]) p[++ptot]=i, miu[i]=-1; for(int j=1, x; j <= ptot && (x=i*p[j]) < Maxn; j++){ p[x]=1; if(i%p[j] == 0) break; miu[x]=-miu[i]; } } for(int i=1; i < Maxn; i++) for(int l=1, r=0; r < i; l=r+1) r=i/(i/l), sum[i]+=1ll*(r-l+1)*(i/l); for(int i=1; i < Maxn; i++) miu[i]+=miu[i-1];}void solve(){ int T=read(); init(); while(T--){ ll n=read(), m=read(), ans=0; if(n > m) swap(n, m); for(int l=1, r=0; r < n; l=r+1){ r=min(n/(n/l), m/(m/l)); ans+=(miu[r]-miu[l-1])*sum[n/l]*sum[m/l]; } printf("%lld\n", ans); }}