[bzoj] 2226 LCMSum || 欧拉函数

原题

求\(\sum^n_{i=1}lcm(i,n)\)

---

\(\sum^n_{i=1}lcm(i,n)\)
\(=\sum^n_{i=1}\frac{n*i}{gcd(i,n)}\)
\(=n*\sum_{d|n}\sum_{i\in[1,n],gcd(i,n)=d}\frac{i}{d}\)
\(=n*\sum_{d|n}\sum_{i\in[1,n/d],gcd(i,n/d)=1}i\)

根据欧拉函数的性质，n>1时，\(\sum_{i\in[1,n]gcd(i,n)=1}i=\frac{\phi(n)*n}{2}\)

所以答案为\(n*\sum_{d|n}\frac{\phi(d)*d}{2}\)
预处理一个欧拉函数就好了（不然会TLE哦）

#include<cstdio>
#include<cmath>
#define N 1000010
typedef long long ll;
using namespace std;
int t,n,m,phi[N],p[N];
bool f[N];

ll read()
{
    ll ans=0,fu=1;
    char j=getchar();
    for (;j<'0' || j>'9';j=getchar()) if (j=='-') fu=-1;
    for (;j>='0' && j<='9';j=getchar()) ans*=10,ans+=j-'0';
    return ans*fu;
}

void init(ll x)
{
    phi[1]=1;
    for (int i=2;i<=x;i++)
    {
	if (!f[i]) phi[i]=i-1,p[++p[0]]=i;
	for (int j=1;j<=p[0] && p[j]*i<=x;j++)
	{
	    f[p[j]*i]=1;
	    if (i%p[j]==0)
	    {
		phi[i*p[j]]=phi[i]*p[j];
		break;
	    }
	    else phi[i*p[j]]=phi[i]*phi[p[j]];
	}
    }
}

ll work(int x) { if (x==1) return 1LL; return (ll)x*phi[x]/2; }

int main()
{
    init(1000000);
    t=read();
    while(t--)
    {
	n=read();
	m=sqrt(n);
	ll ans=0;
	for (int i=1;i<=m;i++)
	    if (n%i==0)
		ans+=work(i),ans+=work(n/i);
	if (m*m==n) ans-=work(m);
	printf("%lld\n",ans*n);
    }
    return 0;
}