SDOI2018 旧试题

题目大意：
求

题解：
大毒瘤。

先反演：

然后套路处理后面那三坨为fa,fb,fc。

为了速度，我们可以直接枚举lcm，将μi!=0&&μj!=0之间建一条边权为lcm(i,j)的边。

如果(u,v,w)这个三元组合法的话：

• 三者互不相同。此时应形成一个三元环。
• 两项相同。此时有一条边即可。
• 三项相同。此时枚举所有的点即可。

代码：

#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 100050
#define MOD 1000000007
#define ll long long
ll pri[N],cnt,mu[N],ft[N][10],mem[N],T,a,b,c;
bool vis[N];
ll fa[N],fb[N],fc[N];
void init()
{
    mu[1]=1;
    for(register ll i=2;i<=100000;i++)
    {
        if(!vis[i])
        {
            mu[i]=-1;
            pri[++cnt]=i;
        }
        for(register ll j=1;i*pri[j]<=100000;j++)
        {
            vis[i*pri[j]]=1;
            if(i%pri[j])mu[i*pri[j]]=-mu[i];
            else break;
        }
    }
    for(register ll i=1;i<=cnt;i++)
    {
        for(register ll j=1;j*pri[i]<=100000;j++)
        {
            mem[pri[i]*j]++;
            ft[pri[i]*j][mem[pri[i]*j]]=pri[i];
        }
    }
}
struct Pair
{
    ll x,v;
    Pair(){}
    Pair(ll x,ll v):x(x),v(v){};
};
vector<Pair>ve[N];
ll d[N];
void initt()
{
    for(register ll i=1;i<=a;i++)fa[i]=0;
    for(register ll i=1;i<=b;i++)fb[i]=0;
    for(register ll i=1;i<=c;i++)fc[i]=0;
    for(register ll i=1;i<=c;i++)
    {
        vector<Pair>tmp;
        swap(ve[i],tmp);
        d[i]=0;
    }
}
ll hs[N];
struct EG
{
    ll f,t,v;
}e[10*N];
ll sol()
{
    if(a>b)swap(a,b);
    if(a>c)swap(a,c);
    if(b>c)swap(b,c);
    for(register ll i=1;i<=a;i++)
        for(register ll j=1;i*j<=a;j++)
        {
            fa[i]+=(a/(i*j));
            if(fa[i]>MOD)fa[i]-=MOD;
        }
    for(register ll i=1;i<=b;i++)
        for(register ll j=1;i*j<=b;j++)
        {
            fb[i]+=(b/(i*j));
            if(fb[i]>MOD)fb[i]-=MOD;
        }
    for(register ll i=1;i<=c;i++)
        for(register ll j=1;i*j<=c;j++)
        {
            fc[i]+=(c/(i*j));
            if(fc[i]>MOD)fc[i]-=MOD;
        }
    ll ct=0;
    for(register ll i=1;i<=c;i++)//lcm
    {
        if(!mu[i])continue;
        for(register ll j=0;j<(1<<mem[i]);j++)
        {
            ll u = 1;
            for(register ll k=1;k<=mem[i];k++)
                if((j>>(k-1))&1)
                    u*=ft[i][k];
            for(register ll k = j;;k=(k-1)&j)
            {
                ll v = i/u;
                for(register ll w=1;w<=mem[i];w++)
                    if((k>>(w-1))&1)
                        v*=ft[i][w];
                if(u>v)
                {
                    d[u]++,d[v]++,ct++;
                    e[ct].f=u,e[ct].t=v,e[ct].v=i;
                }
                if(!k)break;
            }
        }
    }
    for(register ll i=1;i<=ct;i++)
    {
        if(d[e[i].f]<d[e[i].t])
            swap(e[i].f,e[i].t);
        ve[e[i].f].push_back(Pair(e[i].t,e[i].v));
    }
    ll ret = 0;
    ll p1,p2,p3,v1,v2,v3,kk;
    for(register ll i=1;i<=b;i++)
    {
        if(!mu[i])continue;
        //abc
        for(register ll j=0;j<ve[i].size();j++)
            hs[ve[i][j].x]=ve[i][j].v;
        for(register ll j=0;j<ve[i].size();j++)
        {
            for(register ll k=0;k<ve[ve[i][j].x].size();k++)
            {
                if(hs[ve[ve[i][j].x][k].x]&&ve[i][j].x!=ve[ve[i][j].x][k].x)
                {
                    p1=i,p2=ve[i][j].x,p3=ve[p2][k].x;
                    v1 = ve[p1][j].v;v2 = ve[p2][k].v;v3 = hs[p3];
                    kk = mu[p1]*mu[p2]*mu[p3];
                    ret = (ret+kk*fa[v1]*fb[v2]*fc[v3])%MOD;
                    ret = (ret+kk*fa[v1]*fb[v3]*fc[v2])%MOD;
                    ret = (ret+kk*fa[v2]*fb[v1]*fc[v3])%MOD;
                    if(ret<0)ret+=MOD;
                    ret = (ret+kk*fa[v2]*fb[v3]*fc[v1])%MOD;
                    ret = (ret+kk*fa[v3]*fb[v1]*fc[v2])%MOD;
                    ret = (ret+kk*fa[v3]*fb[v2]*fc[v1])%MOD;
                    if(ret<0)ret+=MOD;
                }
            }
        }
        for(register ll j=0;j<ve[i].size();j++)
            hs[ve[i][j].x]=0;
        //aab abb
        for(register ll j=0;j<ve[i].size();j++)
        {
            ll to = ve[i][j].x;
            kk=mu[i];
            p1=i,p2=to,p3=p2;
            v1 = ve[p1][j].v;v2 = p2;v3 = v1;
            ret = (ret+kk*fa[v3]*fb[v1]*fc[v2])%MOD;
            ret = (ret+kk*fa[v2]*fb[v3]*fc[v1])%MOD;
            ret = (ret+kk*fa[v1]*fb[v2]*fc[v3])%MOD;
            if(ret<0)ret+=MOD;
            kk=mu[to];
            p1=i,p2=to,p3=p1;
            v1 = ve[p1][j].v;v2 = v1;v3 = p1;
            ret = (ret+kk*fa[v3]*fb[v1]*fc[v2])%MOD;
            ret = (ret+kk*fa[v2]*fb[v3]*fc[v1])%MOD;
            ret = (ret+kk*fa[v1]*fb[v2]*fc[v3])%MOD;
            if(ret<0)ret+=MOD;
        }
        //aaa
        ret = (ret+mu[i]*fa[i]*fb[i]*fc[i])%MOD;
        if(ret<0)ret+=MOD;
    }
    return ret;
}
int main()
{
    init();
    scanf("%lld",&T);
    while(T--)
    {
        scanf("%lld%lld%lld",&a,&b,&c);
        printf("%lld
",sol());
        initt();
    }
    return 0;
}