题目大意
\(t\)组询问, 每组询问给定\(n\)，求\(\sum_{k=1}^n[n,k]\)，其中\([a,b]\)表示\(a\)和\(b\)的最小公倍数 .
\(t\leq 300000,n\leq 1000000\)
题解
\[\begin{align}
\sum_{k=1}^n[k,n]&=n\sum_{k=1}^n\frac{k}{(k,n)}\\
&=n\sum_{p|n}\frac{1}{p}\sum_{k=1}^nk[(k,n)=p]\\
&=n\sum_{p|n}\sum_{k=1}^{\lfloor\frac{n}{p}\rfloor}k[(k,\frac{n}{p})=1]\\
&=n\sum_{p|n}\frac{\frac{n}{p}\phi(\frac{n}{p})+[\frac{n}{p}=1]}{2}\\
&=n\sum_{p|n}\frac{p\phi(p)+[p=1]}{2}
\end{align}
\]
\sum_{k=1}^n[k,n]&=n\sum_{k=1}^n\frac{k}{(k,n)}\\
&=n\sum_{p|n}\frac{1}{p}\sum_{k=1}^nk[(k,n)=p]\\
&=n\sum_{p|n}\sum_{k=1}^{\lfloor\frac{n}{p}\rfloor}k[(k,\frac{n}{p})=1]\\
&=n\sum_{p|n}\frac{\frac{n}{p}\phi(\frac{n}{p})+[\frac{n}{p}=1]}{2}\\
&=n\sum_{p|n}\frac{p\phi(p)+[p=1]}{2}
\end{align}
\]
用线性筛或者其他方法处理出\(\phi\)函数，调和级数的复杂度预处理，每次查表。或者枚举因子。
时间复杂度：\(O(n\log n)\)或\(O(n+t\sqrt{n})\)
代码
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#include<utility>
#include<map>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
int miu[1000010];
int phi[1000010];
int p[1000010];
int b[1000010];
ll e[1000010];
ll f[1000010];
int cnt;
int maxn=1000000;
map<int,ll> d;
void init()
{
memset(b,0,sizeof b);
int i,j;
cnt=0;
for(i=2;i<=maxn;i++)
{
if(!b[i])
{
p[++cnt]=i;
miu[i]=-1;
phi[i]=i-1;
}
for(j=1;j<=cnt&&i*p[j]<=maxn;j++)
{
b[i*p[j]]=1;
if(i%p[j]==0)
{
miu[i*p[j]]=0;
phi[i*p[j]]=phi[i]*p[j];
break;
}
miu[i*p[j]]=-miu[i];
phi[i*p[j]]=phi[i]*phi[p[j]];
}
}
e[1]=1;
for(i=2;i<=maxn;i++)
e[i]=(ll(i)*phi[i])/2;
for(i=1;i<=maxn;i++)
for(j=i;j<=maxn;j+=i)
f[j]+=e[i];
}
void solve()
{
int n;
scanf("%d",&n);
printf("%lld\n",f[n]*n);
}
int main()
{
// freopen("a.in","r",stdin);
// freopen("a.out","w",stdout);
init();
int t;
scanf("%d",&t);
while(t--)
solve();
return 0;
}