51nod1239欧拉函数求和

求前n个欧拉函数和，n<=10^10
关于积性函数的前缀和见糖老师博客
http://blog.youkuaiyun.com/skywalkert/article/details/50500009
预处理出前500w个欧拉函数前缀和后还需要线性筛一下。用map记录一下搜过的值（可认为记忆化搜索）
代码如下：

#include<bits/stdc++.h>
using namespace std;
#define LL __int64
#define maxn 5000007
#define mod 1000000007
#define ni 500000004
map<LL,LL>mp;
bool isprime[maxn+5];
int prime[maxn+5];
int phi[maxn+5];
LL sum[maxn+5];
int cnt;
void Eular()
{
    cnt=0;
    phi[1]=1;
    for(int i=2; i<=maxn; i++)
    {
        if(!isprime[i])
        {
            phi[i]=i-1;
            prime[++cnt]=i;
        }
        for(int j=1; j<=cnt; j++)
        {
            if(i*prime[j]>maxn)
                break;
            isprime[i*prime[j]]=true;
            if(i%prime[j]==0)
            {
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }
            else
            {
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
    sum[0]=0;
    for(int i=1; i<=maxn; i++)
    {
        sum[i]=(sum[i-1]+phi[i])%mod;
    }
}
LL get_sum(LL m)
{
    if(!m)return 0;
    if(m<maxn)return sum[m];
    if(mp.count(m))return mp[m];
    LL &ans=mp[m];
    for (LL i = 2, j ; i <= m ; i = j + 1LL )
    {
        j = m / ( m / i ) ;
        ans = ( ans + 1LL * ( j - i + 1LL )%mod* get_sum(m / i)%mod ) % mod ;
    }
    return ans = (  m%mod * ( m%mod+ 1LL )%mod *ni%mod - ans + mod ) % mod ;
}
int main()
{
    Eular();
    LL n;
    while(scanf("%I64d",&n)!=EOF)
    {
        printf("%I64d\n",get_sum(n));
    }
    return 0;
}