BZOJ 2401 陶陶的难题I 数论

题目大意：求$\sum_{i=1}^N\sum_{j=1}^NLcm(i,j)$
一开始写了个莫比乌斯反演结果T到死。。。
$\sum_{i=1}^N\sum_{j=1}^NLcm(i,j)\\=\sum_{i=1}^Ni+2\sum_{i=1}^N\sum_{j=1}^{i-1}Lcm(i,j)\\=\sum_{i=1}^Ni+2\sum_{i=1}^N\sum_{j=1}^{i-1}\frac{i*j}{Gcd(i,j)}\\=\sum_{i=1}^Ni+2\sum_{d=1}^N\sum_{i=2}^{\lfloor \frac N d\rfloor}\sum_{j=1}^{i-1}[gcd(i,j)=1]d*i*j\\=\sum_{i=1}^Ni+2\sum_{d=1}^N\sum_{i=2}^{\lfloor \frac N d\rfloor}d*i*\frac{i*\varphi(i)}{2}\\=\sum_{i=1}^Ni+\sum_{d=1}^N\sum_{i=2}^{\lfloor \frac N d\rfloor}d*i^2*\varphi(i)$
然后O(nlogn)计算每一项，求前缀和即可
时间复杂度O(nlogn+T)
WQNMLGB的取个模能死？

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define M 1001001
using namespace std;

struct Int128{
    #define BASE 1000000000000000000ll
    long long num[2];
    void Standardization()
    {
        if(num[0]>=BASE)
            num[0]-=BASE,num[1]++;
    }
    Int128() {}
    Int128(long long _)
    {
        num[0]=_;
        num[1]=0;
        Standardization();
    }
    Int128& operator += (const Int128 &x)
    {
        num[0]+=x.num[0];
        num[1]+=x.num[1];
        Standardization();
        return *this;
    }
    friend ostream& operator << (ostream &_,const Int128 &x)
    {
        if(x.num[1])
        {
            printf("%lld%09d%09d\n",x.num[1],(int)(x.num[0]/1000000000),(int)(x.num[0]%1000000000) );
            return _;
        }
        printf("%lld\n",x.num[0]);
        return _;
    }
};

int phi[M],prime[100100],tot;
bool not_prime[M];
Int128 sum[M];

int n;

void Linear_Shaker()
{
    int i,j;
    for(i=2;i<=1000000;i++)
    {
        if(!not_prime[i])
        {
            phi[i]=i-1;
            prime[++tot]=i;
        }
        for(j=1;prime[j]*i<=1000000;j++)
        {
            not_prime[prime[j]*i]=true;
            if(i%prime[j]==0)
            {
                phi[prime[j]*i]=phi[i]*prime[j];
                break;
            }
            phi[prime[j]*i]=phi[i]*(prime[j]-1);
        }
    }
}

void Pretreatment()
{
    int i,j;
    Linear_Shaker();
    for(i=1;i<=1000000;i++)
    {
        sum[i]+=i;
        for(j=2;i*j<=1000000;j++) 
            sum[i*j]+=Int128((long long)i*j*j*phi[j]);
    }
    for(i=1;i<=1000000;i++)
        sum[i]+=sum[i-1];
}

int main()
{
    int T;
    Pretreatment();
    for(cin>>T;T;T--)
    {
        scanf("%d",&n);
        cout<<sum[n];
    }
    return 0;
}