BZOJ 2154/2693 Crash的数字表格/jzptab (莫比乌斯反演)

题目大意：求$\sum_{i=1}^{n}\sum_{j=1}^{m}lcm(i,j)$的和

易得$\sum_{i=1}^{n}\sum_{j=1}^{m}\frac{ij}{gcd(i,j)}$

套路1：提取出$gcd$

$\sum_{k=1}^{n}\frac{1}{k}\sum_{i=1}^{n}\sum_{j=1}^{m}ij[gcd(i,j)==k]$

$\sum_{k=1}^{n}k\sum_{i=1}^{\left \lfloor \frac{n}{k} \right \rfloor}\sum_{j=1}^{\left \lfloor \frac{m}{k} \right \rfloor}ij[gcd(i,j)==1]$

$\sum_{k=1}^{n}k\sum_{i=1}^{\left \lfloor \frac{n}{k} \right \rfloor}\sum_{j=1}^{\left \lfloor \frac{m}{k} \right \rfloor}ij\sum_{d|gcd(i,j)}\mu(d)$

套路2：继续提取$gcd$

$\sum_{k=1}^{n}k\sum_{d=1}^{\left \lfloor \frac{n}{k} \right \rfloor}\mu(d)\sum_{i=1}^{\left \lfloor \frac{n}{k} \right \rfloor}\sum_{j=1}^{\left \lfloor \frac{m}{k} \right \rfloor}[gcd(i,j)==d]ij$

$\sum_{k=1}^{n}k\sum_{d=1}^{\left \lfloor \frac{n}{k} \right \rfloor}d^{2}\mu(d)\sum_{i=1}^{\left \lfloor \frac{n}{kd} \right \rfloor}\sum_{j=1}^{\left \lfloor \frac{m}{kd} \right \rfloor}ij$

$\sum_{i=1}^{\left \lfloor \frac{n}{kd} \right \rfloor}\sum_{j=1}^{\left \lfloor \frac{m}{kd} \right \rfloor}ij$可以$O(1)$计算出来

套路3：令$Q=kd$

$\sum_{Q=1}^{n}\sum_{i=1}^{\left \lfloor \frac{n}{Q} \right \rfloor}\sum_{j=1}^{\left \lfloor \frac{m}{Q} \right \rfloor}ij\sum_{d|Q}\frac{Q}{d}(d)^{2}\mu(d)$

$\sum_{d|Q}\frac{Q}{d}(d)^{2}\mu(d)$显然是积性函数

然后问题就解决了

#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 10010000
#define maxn 10000000
#define ll long long 
#define uint unsigned int
#define rint register int
using namespace std;

int n,m,T,cnt;
int mu[N],pr[N],use[N];
int f[N],F[N];
const int mod=100000009;

void Pre()
{
    mu[1]=1;f[1]=1;
    for(int i=2;i<=maxn;i++)
    {
        if(!use[i]) pr[++cnt]=i,mu[i]=-1,f[i]=(1ll*i*(1-i))%mod;
        for(rint j=1;j<=cnt&&i*pr[j]<=maxn;j++){
            use[i*pr[j]]=1;
            if(i%pr[j]){
                mu[i*pr[j]]=-mu[i];
                f[i*pr[j]]=1ll*f[i]*f[pr[j]]%mod;
            }else{
                mu[i*pr[j]]=0;
                f[i*pr[j]]=1ll*f[i]*pr[j]%mod;
                break;
            }
        }
    }
    for(int i=1;i-4<=maxn;i+=4)
    {
        F[i+0]=(1ll*F[i-1]+f[i+0])%mod;
        F[i+1]=(1ll*F[i+0]+f[i+1])%mod;
        F[i+2]=(1ll*F[i+1]+f[i+2])%mod;
        F[i+3]=(1ll*F[i+2]+f[i+3])%mod;
    }
}
ll solve(int n,int m)
{
    ll ans=0,sn,sm;
    if(n>m) swap(n,m);
    for(int i=1,la;i<=n;i=la+1)
    {
        la=min(n/(n/i),m/(m/i));
        sn=(1ll*(n/i)*(n/i+1)/2)%mod;
        sm=(1ll*(m/i)*(m/i+1)/2)%mod;
        (ans+=1ll*sn*sm%mod*(F[la]-F[i-1]))%=mod;
    }ans=(ans%mod+mod)%mod;
    return ans;
}

int main()
{
    //freopen("t1.in","r",stdin);
    scanf("%d",&T);
    Pre();
    int n,m;
    while(T--)
    {
        scanf("%d%d",&n,&m);
        printf("%lld\n",solve(n,m));
    }
    return 0;
}

调试部分的代码..留个纪念吧

int ps[N],son[N],d[N],num,nson;
void dfs_son(int s,int dep)
{
    if(dep>num) {son[++nson]=s;return;}
    for(int j=0;j<=d[dep];j++)
        dfs_son(s,dep+1),s*=ps[dep];
}
ll g[N];
void check()
{
    for(int i=1;i<=maxn;i++){
        int sq=sqrt(i),x=i;
        num=nson=0;
        for(int j=1;j<=cnt&&pr[j]<=sq;j++)
            if(x%pr[j]==0){
                ps[++num]=pr[j];
                while(x%pr[j]==0) d[num]++,x/=pr[j];
            }
        if(x!=1) ps[++num]=x,d[num]=1;
        dfs_son(1,1);
        for(int j=1;j<=nson;j++)
            (g[i]+=mu[son[j]]*son[j]%mod)%=mod,
            son[j]=0;
        g[i]=g[i]*i%mod;
        for(int i=1;i<=num;i++) 
            d[i]=0;
    }
}