bzoj 2154: Crash的数字表格 莫比乌斯反演

题意

求 ∑ni=1∑mj=1lcm(i,j)
n,m<=10000000

分析

题解见Po神的论文
或者看这篇比较详细的博客
恩全OJ倒数第二……

代码

#include
#include
#include
#include
#include
#define MOD 20101009
#define ll long long
using namespace std;

const int MAXN=10000000;
const int N=10000005;

int prime[1000005],tot,mu[N],n,m;
ll s[N];
bool not_prime[N];

void getmu(int n)
{
    mu[1]=1;
    for (int i=2;i<=n;i++)
    {
        if (!not_prime[i])
        {
            prime[++tot]=i;
            mu[i]=-1;
        }
        for (int j=1;j<=tot&&i*prime[j]<=n;j++)
        {
            not_prime[i*prime[j]]=1;
            if (i%prime[j]==0)
            {
                mu[i*prime[j]]=0;
                break;
            }
            mu[i*prime[j]]=-mu[i];
        }
    }
    for (int i=1;i<=n;i++)
        s[i]=(s[i-1]+(ll)mu[i]*i*i%MOD)%MOD;
}

ll getsum(int n)
{
    return (ll)n*(n+1)%MOD*10050505%MOD;
}

ll F(int n,int m)
{
    if (n>m) swap(n,m);
    ll ans=0;
    for (int i=1,last;i<=n;i=last+1)
    {
        last=min(n/(n/i),m/(m/i));
        ans=(ans+getsum(n/i)*getsum(m/i)%MOD*(s[last]-s[i-1])%MOD)%MOD;
    }
    return ans;
}

int main()
{
    getmu(MAXN);
    scanf("%d%d",&n,&m);
    if (n>m) swap(n,m);
    ll ans=0;
    for (int i=1,last;i<=n;i=last+1)
    {
        last=min(n/(n/i),m/(m/i));
        ans=(ans+F(n/i,m/i)*(getsum(last)-getsum(i-1))%MOD)%MOD;
    }
    printf("%lld",(ans+MOD)%MOD);
    return 0;
}