【BZOJ 2818】Gcd

这题一开始我竟然想要用与能量采集差不多的思路去做= =（no zuo no die,why you try?）
有个显然的转化
$\sum_{x=1}^{n}\sum_{y=1}^{n} [gcd(x,y)==P]= \sum_{x=1}^{\lfloor n/P \rfloor}\sum_{y=1}^{\lfloor n/P \rfloor}[gcd(x,y)==1]=2*\sum_{i=2}^{\lfloor n/P \rfloor} \phi (i)+1$
然后我们线性筛出欧拉函数，在前缀和就可以$O(n)$求出答案了
code：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
long long f[10000001];
int phi[10000001],p[700001];
long long ans=0;
long long s[10000001];
int a[10000001],n;
int main()
{
    long long i,t=0,j;
    scanf("%lld",&n);
    a[1]=1; phi[1]=1;
    for (i=2;i<=n;++i)
      {
        if (!a[i])
          {
            p[++t]=i;
            phi[i]=i-1;
          }
        for (j=1;j<=t;++j)
          {
            if (i*p[j]>n) break;
            a[i*p[j]]=1;
            if (i%p[j]==0)
              {
                phi[i*p[j]]=phi[i]*p[j];
                break;
              }
            else
              phi[i*p[j]]=phi[i]*(p[j]-1);
          }
      }
    s[1]=0;
    for (i=2;i<=n;++i) s[i]=s[i-1]+phi[i];
    for (i=1;i<=n;++i) f[i]=2*s[i]+1;
    for (i=1;i<=t;++i) ans=ans+f[n/p[i]];
    printf("%lld\n",ans);
}