[egin{eqnarray*}
ans&=&sum_{i=1}^ng(i)\
&=&sum_{i=1}^nsum_{d|i}mu^2(d)\
&=&sum_{i=1}^nsum_{d|i}sum_{k^2|d}mu(k)\
&=&sum_{k=1}^nmu(k)sum_{k^2|d}lfloorfrac{n}{d}
floor\
&=&sum_{k=1}^nmu(k)sum_{i=1}^{lfloorfrac{n}{k^2}
floor}lfloorfrac{n}{k^2i}
floor\
&=&sum_{k=1}^{sqrt{n}}mu(k)S(lfloorfrac{n}{k^2}
floor)
end{eqnarray*}]
其中
[S(n)=sum_{i=1}^nlfloorfrac{n}{i} floor]
枚举所有$k$，然后分段计算$S$即可，当$n$比较小的时候可以记忆化$S(n)$。
时间复杂度
[egin{eqnarray*}
T(n)&=&O(sqrt{n}+sum_{i=1}^{sqrt{n}}sqrt{frac{n}{i^2}})\
&=&O(sqrt{n}sum_{i=1}^{sqrt{n}}frac{1}{i})\
&=&O(sqrt{n}log n)
end{eqnarray*}]
#include<cstdio>
typedef long long ll;
const int N=1000010,P=1000000007;
int T,C,tot,p[N/10],i,j,ans,f[N];char mu[N],v[N];ll n;
inline int F(ll n){
if(n<N)if(f[n])return f[n];
ll t=0;
for(ll i=1,j;i<=n;i=j+1)j=n/(n/i),t+=n/i*(j-i+1);
t%=P;
if(n<N)f[n]=t;
return t;
}
int main(){
for(mu[1]=1,i=2;i<N;i++){
if(!v[i])mu[i]=-1,p[tot++]=i;
for(j=0;j<tot&&i*p[j]<N;j++){
v[i*p[j]]=1;
if(i%p[j])mu[i*p[j]]=-mu[i];else break;
}
}
for(scanf("%d",&T);T--;printf("Case #%d: %d
",++C,(ans+P)%P)){
scanf("%I64d",&n);
for(ans=0,i=1;i<=n/i;i++)if(mu[i])ans=(ans+F(n/i/i)*mu[i])%P;
}
return 0;
}