BZOJ 4804: 欧拉心算

数论题不多BB，直接开始推导吧：

(sum_{i=1}^n sum_{j=1}^n phi(gcd(i,j)))

(=sum_{i=1}^n sum_{j=1}^n sum_{d=1}^n [gcd(i,j)=d]phi(d))

(=sum_{d=1}^n phi(d)cdot(sum_{i=1}^{lfloorfrac{n}{d} floor} sum_{j=1}^{lfloorfrac{n}{d} floor} [gcd(i,j)=1]))

然后(sum_{i=1}^{lfloorfrac{n}{d} floor} sum_{j=1}^{lfloorfrac{n}{d} floor} [gcd(i,j)=1]=2cdotsum_{i=1}^{lfloorfrac{n}{d} floor} phi(i) -1)，这个也是一个比较常用的结论了吧。具体可以看Luogu P2568 GCD的推导，之后为了方便，我们记(sum_x=sum_{i=1}^x phi(i))

则上式(=sum_{d=1}^n(phi(d)cdot (2cdot sum_{lfloorfrac{n}{d} floor}-1)))

(=2cdot sum_{d=1}^n(phi(d)cdot sum_{lfloorfrac{n}{d} floor})-sum_n)

接下来就好做了，我们先把欧拉函数求前缀和，然后对于询问除法分块处理不同的(sum_{lfloorfrac{n}{d} floor})即可。

复杂度(O(n+Tsqrt n))，轻松跑过

CODE

#include<cstdio>
#define RI register int
using namespace std;
const int P=1e7;
int prime[P+5],phi[P+5],cnt,n,t; long long ans,sum[P+5]; bool vis[P+5];
inline void resolve(void)
{
    vis[1]=sum[1]=phi[1]=1; for (RI i=2;i<=P;++i)
    {
        if (!vis[i]) prime[++cnt]=i,phi[i]=i-1;
        for (RI j=1;j<=cnt&&i*prime[j]<=P;++j)
        {
            vis[i*prime[j]]=1; if (i%prime[j]) phi[i*prime[j]]=phi[i]*(prime[j]-1);
            else { phi[i*prime[j]]=phi[i]*prime[j]; break; }
        }
        sum[i]=sum[i-1]+phi[i];
    }
}
int main()
{
    for (scanf("%d",&t),resolve();t;--t)
    {
        scanf("%d",&n); ans=0; RI i,l,r;
        for (l=1;l<=n;l=r+1) r=n/(n/l),ans+=sum[n/l]*(sum[r]-sum[l-1]);
        printf("%lld
",(ans<<1)-sum[n]);
    }
    return 0;
}