推狮子的部分
[sum_{i=1}^Asum_{j=1}^Bsum_{k=1}^Csigma(ijk)
=sum_{i=1}^Asum_{j=1}^Bsum_{k=1}^Csum_{x|i}sum_{y|j}sum_{z|k}epsilon(gcd(x,y))epsilon(gcd(y,z))epsilon(gcd(x,z))\
=sum_{i=1}^Asum_{x|i}sum_{j=1}^Bsum_{y|j}sum_{k=1}^Csum_{z|k}epsilon(gcd(x,y))epsilon(gcd(y,z))epsilon(gcd(x,z))\
=sum_{x=1}^Alfloorfrac{A}{x}
floorsum_{y=1}^Blfloorfrac{B}{y}
floorsum_{z=1}^Clfloorfrac{C}{z}
floorepsilon(gcd(x,y))epsilon(gcd(y,z))epsilon(gcd(x,z))\
=sum_{x=1}^Alfloorfrac{A}{x}
floorsum_{y=1}^Blfloorfrac{B}{y}
floorsum_{z=1}^Clfloorfrac{C}{z}
floorsum_{d|x,d|y}mu(d)sum_{p|y,p|z}mu(p)sum_{q|x,q|z}mu(q)\
=sum_{d=1}^{min(A,B)}mu(d)sum_{p=1}^{min(B,C)}mu(p)sum_{q=1}^{min(A,C)}mu(q)sum_{d|x,q|x}^Alfloorfrac{A}{x}
floorsum_{d|y,p|y}^Blfloorfrac{B}{y}
floorsum_{p|z,q|z}^Clfloorfrac{C}{z}
floor\
=sum_{d=1}^{min(A,B)}mu(d)sum_{p=1}^{min(B,C)}mu(p)sum_{q=1}^{min(A,C)}mu(q)sum_{lcm(d,q)|x}^Alfloorfrac{A}{x}
floorsum_{lcm(d,p)|y}^Blfloorfrac{B}{y}
floorsum_{lcm(p,q)|z}^Clfloorfrac{C}{z}
floor\
ext{define } f(n,t)=sum_{t|x}lfloorfrac{n}{x}
floor ,N=max(A,B,C)\
cdots=sum_{d=1}^Nmu(d)sum_{p=1}^Nmu(p)sum_{q=1}^Nmu(q)f(A,lcm(d,q))f(B,lcm(d,p))f(C,lcm(p,q))\
]
计算答案
其中(f(n,t))可以(O(nlog n))预处理。
考虑对(T)个点连边建图，(u)、(v)之间有边当且仅当(mu(u) ot=0,mu(v) ot=0,lcm(a,b) ot>T)。那么图中的每个三元环都能算入答案，这里的三元环还包括只有俩点的和只有单点的。
对于包含三个点的(<d,p,q>)的贡献为
[mu(d)mu(p)mu(q) imes\
(f(A,lcm(d,q))f(B,lcm(d,p))f(C,lcm(p,q))+\
f(A,lcm(d,q))f(B,lcm(p,q))f(C,lcm(d,p))+\
f(A,lcm(d,p))f(B,lcm(d,q))f(C,lcm(p,q))+\
f(A,lcm(d,p))f(B,lcm(p,q))f(C,lcm(d,q))+\
f(A,lcm(p,q))f(B,lcm(d,p))f(C,lcm(d,q))+\
f(A,lcm(p,q))f(B,lcm(d,q))f(C,lcm(d,p)))
]
对于包含两个、一个的环同理。统计三元环的方法参照不常用的黑科技——「三元环」。
参考实现
#include <bits/stdc++.h>
using namespace std;
const int N=1e5+10;
const int mod=1e9+7;
//[SDOI2018]旧试题
int pr[N],cnt;
int mu[N],bk[N],deg[N];
long long fa[N],fb[N],fc[N];
bool vis[N];
struct Node {int u,v,w;};
struct Edge {int ver,len;};
Node ech[N*300];
vector<Edge> e[N];
int solve(int a,int b,int c) {
long long ans=0;
int n=max(a,max(b,c)),m=0;
for(int i=1; i<=n; ++i) {
deg[i]=fa[i]=fb[i]=fc[i]=0;
e[i].clear();
}
for(int i=1; i<=n; ++i) {
for(int x=i; x<=n; x+=i) {
fa[i]+=a/x,fb[i]+=b/x,fc[i]+=c/x;
}
}
for(int i=1; i<=a&&i<=b&&i<=c; ++i) {
if(mu[i]) ans+=mu[i]*mu[i]*mu[i]*fa[i]*fb[i]*fc[i];
}
for(int g=1; g<=n; ++g) {
for(int i=1; i*g<=n; ++i) if(mu[i*g]) {
for(int j=i+1; 1LL*i*j*g<=n; ++j) if(mu[j*g]&&__gcd(i,j)==1) {
int u=i*g,v=j*g,w=i*j*g;
deg[u]++,deg[v]++,ech[++m]=(Node){u,v,w};
ans+=mu[u]*mu[u]*mu[v]*(fa[u]*fb[w]*fc[w]+fa[w]*fb[u]*fc[w]+fa[w]*fb[w]*fc[u]);
ans+=mu[u]*mu[v]*mu[v]*(fa[v]*fb[w]*fc[w]+fa[w]*fb[v]*fc[w]+fa[w]*fb[w]*fc[v]);
}
}
}
for(int i=1; i<=m; ++i) {
if(deg[ech[i].u]<deg[ech[i].v]||(deg[ech[i].u]==deg[ech[i].v]&&ech[i].u<ech[i].v))
swap(ech[i].u,ech[i].v);
e[ech[i].u].push_back((Edge){ech[i].v,ech[i].w});
}
#define veit vector<Edge>::iterator
for(int i=1; i<=n; ++i) if(mu[i]) {
for(veit j=e[i].begin(); j!=e[i].end(); ++j) bk[j->ver]=j->len;
for(veit j=e[i].begin(); j!=e[i].end(); ++j) {
for(veit k=e[j->ver].begin(); k!=e[j->ver].end(); ++k) {
if(!bk[k->ver]) continue;
ans+=mu[i]*mu[j->ver]*mu[k->ver]*(
fa[j->len]*fb[k->len]*fc[bk[k->ver]]+fa[j->len]*fb[bk[k->ver]]*fc[k->len]+fa[k->len]*fb[j->len]*fc[bk[k->ver]]+
fa[k->len]*fb[bk[k->ver]]*fc[j->len]+fa[bk[k->ver]]*fb[j->len]*fc[k->len]+fa[bk[k->ver]]*fb[k->len]*fc[j->len]
);
}
}
for(veit j=e[i].begin(); j!=e[i].end(); ++j) bk[j->ver]=0;
}
return (ans%mod+mod)%mod;
}
void sieve() {
mu[1]=1;
for(int i=2; i<N; ++i) {
if(!vis[i]) mu[pr[++cnt]=i]=-1;
for(int j=1; j<=cnt&&i*pr[j]<N; ++j) {
vis[i*pr[j]]=1;
if(i%pr[j]==0) break;
else mu[i*pr[j]]=-mu[i];
}
}
}
int main() {
sieve();
int T,a,b,c;
scanf("%d",&T);
while(T--) {
scanf("%d%d%d",&a,&b,&c);
printf("%d
",solve(a,b,c));
}
return 0;
}