Sol

推导：\(n<m，p为质数\)
\(ans=\sum_p\sum_{i=1}^{\frac{n}{p}}\mu(i)\frac{n}{pi}\frac{m}{pi}\)
\(=\sum_{k=1}^{n}\frac{n}{k}\frac{m}{k}\sum_{p|k}\mu(\frac{k}{p})\)
\(\sum_{p|k}\mu(\frac{k}{p})\)可以暴力预处理，也可以在筛的时候计算出

暴力求

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Zsydalao 666
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e7 + 1);IL ll Read(){char c = '%'; ll x = 0, z = 1;for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';return x * z;
}int prime[_], num, mu[_], f[_];
bool isprime[_];IL void Prepare(){isprime[1] = 1; mu[1] = 1;for(RG int i = 2; i < _; ++i){if(!isprime[i]) prime[++num] = i, mu[i] = -1;for(RG int j = 1; j <= num && i * prime[j] < _; ++j){isprime[i * prime[j]] = 1;if(i % prime[j])  mu[i * prime[j]] = -mu[i];else{  mu[i * prime[j]] = 0; break;  }}}for(RG int i = 1; i < _; ++i)for(RG int j = 1; j <= num && i * prime[j] < _; ++j)f[i * prime[j]] += mu[i];for(RG int i = 1; i < _; ++i) f[i] += f[i - 1];
}int main(RG int argc, RG char *argv[]){Prepare();for(RG int T = Read(); T; --T){RG ll n = Read(), m = Read(), ans = 0;if(n > m) swap(n, m);for(RG ll k = 1, j; k <= n; k = j + 1){j = min(n / (n / k), m / (m / k));ans += (n / k) * (m / k) * (f[j] - f[k - 1]);}printf("%lld\n", ans);}return 0;
}

筛的时候处理

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Zsydalao 666
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e7 + 1);IL ll Read(){char c = '%'; ll x = 0, z = 1;for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';return x * z;
}int prime[_], num, mu[_], f[_];
bool isprime[_];IL void Prepare(){isprime[1] = 1; mu[1] = 1;for(RG int i = 2; i < _; ++i){if(!isprime[i]) prime[++num] = i, mu[i] = -1, f[i] = 1;for(RG int j = 1; j <= num && i * prime[j] < _; ++j){isprime[i * prime[j]] = 1;if(i % prime[j])  mu[i * prime[j]] = -mu[i], f[i * prime[j]] = mu[i] - f[i];else{  mu[i * prime[j]] = 0; f[i * prime[j]] = mu[i]; break;  }}}for(RG int i = 1; i < _; ++i) f[i] += f[i - 1];
}int main(RG int argc, RG char *argv[]){Prepare();for(RG int T = Read(); T; --T){RG ll n = Read(), m = Read(), ans = 0;if(n > m) swap(n, m);for(RG ll k = 1, j; k <= n; k = j + 1){j = min(n / (n / k), m / (m / k));ans += (n / k) * (m / k) * (f[j] - f[k - 1]);}printf("%lld\n", ans);}return 0;
}