Mobius first channel inversion. . . $ Qwq $
设$f(d)=\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)==d]$
$F(n)=\sum_{n|d}f(d)=\lfloor \frac{N}{n} \rfloor \lfloor \frac{M}{n} \rfloor$
$f(n)=\sum_{n|d}\mu(\frac{d}{n})F(d)$
$ Ans = \ sum_ {p \} In the first F (P) $
$=\sum_{p\in pri}\sum_{p|d}\mu(\frac{d}{p})F(d)$
Provided $ d '= d / p $
$=\sum_{p\in pri}\sum_{d'}^{min(\lfloor \frac{N}{p} \rfloor,\lfloor \frac{M}{p} \rfloor)} \mu(d') F(d'p)$
$=\sum_{p\in pri}\sum_{d'}^{min(\lfloor \frac{N}{p} \rfloor,\lfloor \frac{M}{p} \rfloor)} \mu(d') \lfloor \frac{N}{d'p} \rfloor \lfloor \frac{M}{d'p} \rfloor$
Set $ T = d'p $
$ = \ Sum_ {T = 1} ^ {min (N, M)} \ sum_ {t | T, t \ in pri} \ mu (\ frac {T} {t}) \ lfloor \ frac {N} { T} \ rfloor \ lfloor \ frac {M} {T} \ rfloor $ (in fact, the previous paragraph is removed for better understanding $ \ $ SUM process)
$=\sum_{T=1}^{min(N,M)}\lfloor \frac{N}{T} \rfloor \lfloor \frac{M}{T} \rfloor \sum_{t|T,t\in pri} \mu(\frac{T}{t})$
For $ \ lfloor \ frac {N} {T} \ rfloor \ lfloor \ frac {M} {T} \ rfloor $ a divisible block for $ a (T) = \ sum_ {t | T, t \ in pri } \ mu (\ frac {T} {t}) $ Å sieve with a similar idea to $ a (T) $ sieve out and then make a prefix. .
#include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<cctype> #include<cstdlib> #include<vector> #include<queue> #include<map> #include<set> #define ll long long #define R register int using namespace std; namespace Fread { static char B[1<<15],*S=B,*D=B; #define getchar() (S==D&&(D=(S=B)+fread(B,1,1<<15,stdin),S==D)?EOF:*S++) inline int g() { R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-1:fix; do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix; } }using Fread::g; int t,n,m,cnt,pri[1000000],a[10000001],mu[10000001]; bool v[10000001]; long long sum[10000001]; inline void MU(int n) { mu[1]=1; for(R i=2;i<=n;++i) { if(!v[i]) pri[++cnt]=i,mu[i]=-1; for(R j=1;j<=cnt&&i*pri[j]<=n;++j) { v[i*pri[j]]=true; if(i%pri[j]==0) break; else mu[i*pri[j]]=-mu[i]; } } for(R j=1;j<=cnt;++j) for(R i=1;i*pri[j]<=n;++i) a[i*pri[j]]+=mu[i]; for(R i=1;i<=n;++i) sum[i]=sum[i-1]+(ll)a[i]; } signed main() { #ifdef JACK freopen("NOIPAK++.in","r",stdin); #endif MU(10000000); t=g(); while(t--) { register long long ans=0; n=g(),m=g(); n>m?(void)(swap(n,m)):(void)0; for(R l=1,r;l<=n;l=r+1) { r=min(n/(n/l),m/(m/l)); ans+=(ll)(n/l)*(m/l)*(sum[r]-sum[l-1]); }printf("%lld\n",ans); } }
2019.06.09