$$\begin{align*}
&\ \ \ \ \sum_{k=1}^{n}\sum_{i=1}^{k}\sum_{j=1}^{k}(i,j,k)\\
&=\sum_{p=1}^{n}p\sum_{k=1}^{\lfloor\frac{n}{p}\rfloor}\sum_{i=1}^{k}\sum_{j=1}^{k}[(i,j,k)=1]\\
&=\sum_{p=1}^{n}p\sum_{k=1}^{\lfloor\frac{n}{p}\rfloor}\sum_{i=1}^{k}\sum_{j=1}^{k}\sum_{d|i,d|j,d|k}\mu(d)\\
&=\sum_{p=1}^{n}p\sum_{d=1}^{\lfloor\frac{n}{p}\rfloor}\mu(d)\sum_{k=1}^{\lfloor\frac{n}{pd}\rfloor}\sum_{i=1}^{k}\sum_{j=1}^{k}1\\
&=\sum_{p=1}^{n}p\sum_{d=1}^{\lfloor\frac{n}{p}\rfloor}\mu(d)f(\lfloor\frac{n}{pd}\rfloor)\\
&=\sum_{T=1}^{n}f(\lfloor\frac{n}{T}\rfloor)\sum_{d|T}\mu(d)\cdot\frac{T}{d}\\
&=\sum_{T=1}^{n}f(\lfloor\frac{n}{T}\rfloor)\varphi(T)
\end{align*}$$
其中$f(n)=\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$，最后一步是根据$id*\mu=\varphi$得到的。
观察这个式子发现可以根号加速，$\varphi$的前缀和用杜教筛求出。通过记忆化能做到玄学复杂度。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define rep(i,l,r) for (int i=(l); i<=(r); i++) 5 using namespace std; 6 7 const int N=1000010; 8 int n,mod,m=1000000,inv,b[N],p[N],tot,phi[N],Phi[N],ans; 9 10 int ksm(int a,int b){ 11 int res=1; 12 for (; b; a=1ll*a*a%mod,b>>=1) 13 if (b & 1) res=1ll*res*a%mod; 14 return res; 15 } 16 17 int S(int x){ return 1ll*x*(x+1)%mod*(x+x+1)%mod*inv%mod; } 18 19 void init(int n){ 20 rep(i,2,n){ 21 if (!b[i]) p[++tot]=i,phi[i]=i-1; 22 for (int j=1; i*p[j]<=n; ++j){ 23 b[i*p[j]]=1; 24 if (i%p[j]) phi[i*p[j]]=phi[i]*(p[j]-1); 25 else { phi[i*p[j]]=phi[i]*p[j]; break; } 26 } 27 } 28 } 29 30 int P(int x){ 31 if (x<=m) return phi[x]; 32 if (~Phi[n/x]) return Phi[n/x]; 33 int res=0; 34 for (int i=2,j; i<=x; i=j+1) 35 j=x/(x/i),res=(res+1ll*(j-i+1)*P(x/i))%mod; 36 res=((1ll*x*(x+1)>>1)-res+mod)%mod; 37 return Phi[n/x]=res; 38 } 39 40 int main(){ 41 freopen("a.in","r",stdin); 42 freopen("a.out","w",stdout); 43 scanf("%d%d",&n,&mod); inv=ksm(6,mod-2); phi[1]=1; init(m); 44 rep(i,1,m) phi[i]=(phi[i]+phi[i-1])%mod; 45 memset(Phi,-1,sizeof(Phi)); 46 for (int i=1,j; i<=n; i=j+1) 47 j=n/(n/i),ans=(ans+1ll*(P(j)-P(i-1)+mod)*S(n/i))%mod; 48 printf("%d\n",ans); 49 return 0; 50 }