思路：

莫比乌斯反演+整除分块

代码：

#pragma GCC optimize(2)#pragma GCC optimize(3)#pragma GCC optimize(4)#include
    
     using namespace std;#define y1 y11#define fi first#define se second#define pi acos(-1.0)#define LL long long//#define mp make_pair#define pb emplace_back#define ls rt<<1, l, m#define rs rt<<1|1, m+1, r#define ULL unsigned LL#define pll pair
     
      #define pli pair
      
       #define pii pair
       
        #define piii pair
        
         #define pdd pair
         
          #define mem(a, b) memset(a, b, sizeof(a))#define debug(x) cerr << #x << " = " << x << "\n";#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);//headconst int N = 1e7 + 10;int T, n, m;int prime[N/10], mu[N], sum[N], cnt;bool not_p[N];LL s[N];void seive() { mu[1] = 1; for (int i = 2; i < N; ++i) { if(!not_p[i]) prime[++cnt] = i, mu[i] = -1; for (int j = 1; j <= cnt && i*prime[j] < N; ++j) { not_p[i*prime[j]] = true; if(i%prime[j] == 0) { mu[i*prime[j]] = 0; break; } mu[i*prime[j]] = -mu[i]; } } for (int i = 2; i < N; ++i) { if(!not_p[i]) for (int j = i; j < N; j += i) sum[j] += mu[j/i]; } for (int i = 1; i < N; ++i) s[i] = s[i-1] + sum[i];}int main() { seive(); scanf("%d", &T); while(T--) { scanf("%d %d", &n, &m); LL ans = 0; int up = min(n, m); for (int l = 1, r; l <= up; l = r+1) { r = min(n/(n/l), m/(m/l)); ans += (n/l)*1LL*(m/l)*(s[r]-s[l-1]); } printf("%lld\n", ans); } return 0;}