# [bzoj] 2693 jzptab || 莫比乌斯反演

### 原题

$\sum^{n}_{x=1}\sum^{m}_{y=1}lcm(x,y)$

$\sum^{n}_{x=1}\sum^{m}_{y=1}lcm(x,y)$

$=\sum_{p}\sum^{\lfloor \frac{n}{p} \rfloor}_{x=1}\sum^{\lfloor \frac{m}{p} \rfloor}_{y=1}i*j*p[gcd(x,y)==1]$

$=\sum_{p}\sum^{\lfloor \frac{n}{p} \rfloor}_{x=1}\sum^{\lfloor \frac{m}{p} \rfloor}_{y=1}\sum_{d|gcd(x,y)}\mu(d)*x*y*p$

$=\sum_{p}p*\sum_d\sum^{\lfloor \frac{n}{p} \rfloor}_{d|x}\sum^{\lfloor \frac{m}{p} \rfloor}_{d|y}\sum_{d|gcd(x,y)}\mu(d)*x*y$

$=\sum_{p}p*\sum_d\mu(d)*d^2*\sum^{\lfloor \frac{n}{p} \rfloor}_{d|x}\sum^{\lfloor \frac{m}{p} \rfloor}_{d|y}\frac{x}{d}*\frac{y}{d}$

$sum(i,j)=\sum^{\lfloor \frac{n}{pd} \rfloor}_{i=1}\sum^{\lfloor \frac{m}{pd} \rfloor}_{j=1}i*j=((\lfloor \frac{n}{pd} \rfloor)*(\lfloor \frac{n}{pd} \rfloor+1)/2)*((\lfloor \frac{m}{pd} \rfloor)*(\lfloor \frac{m}{pd} \rfloor+1)/2)$

$=\sum_{p}p*\sum_d\mu(d)*d^2*sum(\lfloor \frac{n}{pd} \rfloor,\lfloor \frac{m}{pd} \rfloor)$

#include<cstdio>
#include<algorithm>
#define N 10000010
#define mod 100000009ll
typedef long long ll;
using namespace std;
ll pri[N>>3],tot,g[N],sum[N],t,n,m,ans;
bool vis[N];

void init()
{
g[1]=sum[1]=1;
for (ll i=2;i<N;i++)
{
if (!vis[i])
{
pri[++tot]=i;
g[i]=((-(ll)(i-1)*i)%mod+mod)%mod;
}
for (ll j=1;j<=tot && pri[j]*i<N;j++)
{
vis[i*pri[j]]=1;
if (i%pri[j]==0)
{
g[i*pri[j]]=g[i]*pri[j]%mod;
break;
}
g[i*pri[j]]=g[i]*g[pri[j]]%mod;
}
sum[i]=sum[i-1]+g[i];
}
}

ll F(ll x,ll y)
{
return ((ll)(x+1)*x/2)%mod*(((ll)(y+1)*y/2)%mod)%mod;
}

int main()
{
init();
scanf("%lld",&t);
while (t--)
{
scanf("%lld%lld",&n,&m);
if (n>m) swap(n,m);
ans=0;
for (int i=1,last=0;i<=n;i=last+1)
{
last=min(n/(n/i),m/(m/i));
ans+=F(n/i,m/i)*(sum[last]-sum[i-1]+mod)%mod;
ans%=mod;
}
printf("%lld\n",ans);
}
return 0;
}

posted @ 2018-01-05 16:26  Mrha  阅读(102)  评论(0编辑  收藏  举报