题目大意：多组数据，每次给定\(a,b,d\)，询问\(\sum_{i=1}^{a}\sum_{j=1}^{b}[gcd(i,j)=d]\)
莫比乌斯反演
解析：
\[\begin{aligned}ans &= \sum_{i=1}^{a}\sum_{j=1}^{b}[gcd(i,j)=d] \\ &=\sum_{i=1}^{\lfloor \frac{a}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{b}{d} \rfloor}[gcd(i,j)=1]\end{aligned}
\]
方便起见，令\(n=\lfloor \frac{a}{d} \rfloor\)，\(m=\lfloor \frac{b}{d} \rfloor\)
\[\begin{aligned}ans &= \sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=1] \\ &= \sum_{i=1}^{n}\sum_{j=1}^{m}\epsilon(gcd(i,j))\end{aligned}
\]
因为\(\mu * 1=\epsilon\)，且\(d \mid gcd(i,j) \Longleftrightarrow d\mid i,d \mid j\)
\[\begin{aligned}ans &= \sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{d\mid i,d \mid j}\mu(d) \\ &= \sum_{d=1}^{min(n,m)}\mu(d)\sum_{i=1}^{n}\sum_{j=1}^{m}[d \mid i][d \mid j] \\ &= \sum_{d=1}^{min(n,m)}\mu(d)\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor \end{aligned}
\]
筛出\(\mu\)，整除分块求解即可
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
const int maxn = 1e5;
int vis[maxn],mu[maxn];
inline int sum(int a,int b){return mu[b] - mu[a - 1];}
vector<int> pri;
inline void sieve(){
mu[1] = 1;
for(int i = 2;i < maxn;i++){
if(!vis[i]){
pri.push_back(i);
mu[i] = -1;
}
for(int x : pri){
if(1ll * x * i >= maxn)break;
vis[i * x] = 1;
if(i % x){
mu[i * x] = mu[i] * mu[x];
}else{
mu[i * x] = 0;
break;
}
}
}
for(int i = 1;i < maxn;i++)mu[i] += mu[i - 1];
}
int t,a,b,n,m,d;
inline void solve(){
cin >> a >> b >> d;
n = a / d,m = b / d;
int ans = 0;
for(int l = 1,r;l <= min(n,m);l = r + 1){
r = min(n / (n / l),m / (m / l));
ans += sum(l,r) * (n / l) * (m / l);
}
cout << ans << '\n';
}
int main(){
sieve();
cin >> t;
while(t--)solve();
return 0;
}