題意：

$d(n)$為$n$的約數個數， 求$\sum_{i = 1} ^ {n}\sum_{j = 1}^{m}\d(i * j)$

思路：
$d(i * j) = \sum_{x|i}\sum_{y|j}[gcd(x, y) = 1]$

$= \sum_{x | i}\sum{y | j}\sum{k | gcd(x, y)}\mu(k)$

$= \sum_{k}\mu(k)\sum_{x | i}\sum_{y | j}[k | x][k | y]$

$= \sum_{k}\mu(k)\sum_{x | \frac{i}{k}}\sum_{y|\frac{j}{k}}$

$= \sum_{k}\mu(k)\d(\frac{i}{k})\d(\frac{j}{k})$

則$\sum_{i = 1} ^ {n}\sum_{j = 1}^{m}\d(i * j)$

$=\sum_{k}\mu(k)\sum_{i = 1} ^ {n}\d(\frac{i}{k})\sum_{j = 1}^{m}\d(\frac{j}{k})$

$=\sum_{k}\mu(k)\sum_{i = 1} ^ {\lfloor\frac{n}{k}\rfloor}\d(i)\sum_{j = 1}^{\lfloor\frac{m}{k}\rfloor}\d(j)$

線性篩求$mu(i)$和$d(i)$的字首和，分塊求解

Code:

#pragma GCC optimize(3)
#pragma GCC optimize(2)
#include <map>
#include <set>
// #include <array>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cstring>
#include <sstream>
#include <iostream>
#include 
 
<stdlib.h>
#include <algorithm>
// #include <unordered_map>

using namespace std;

typedef long long ll;
typedef pair<int, int> PII;

#define Time (double)clock() / CLOCKS_PER_SEC

#define sd(a) scanf("%d", &a)
#define sdd(a, b) scanf("%d%d", &a, &b)
#define slld(a) scanf("%lld", &a)
#define slldd(a, b) scanf("%lld%lld", &a, &b)

const int N = 1e7 + 20;
const int M = 1e5 + 20;
const int mod = 1e9 + 7;
const double eps = 1e-6;

ll cnt = 0, primes[N], phi[N], mu[N], d[N], c[N], sum[N] = {0}, sd[N] = {0};
map<ll, ll> su;
bool st[N];

int n, m, p, k;

void get(ll n){
    // phi[1] = 1;
    mu[1] = 1;
    d[1] = 1;
    for(ll i = 2; i <= n; i ++){
        if(!st[i]){
            primes[cnt ++] = i;
            // phi[i] = i - 1;
            mu[i] = -1;
            d[i] = 2;
            c[i] = 1;
        }
        for(int j = 0; primes[j] <= n / i; j ++){
            st[i * primes[j]] = true;
            if(i % primes[j] == 0){
                // phi[i * primes[j]] = primes[j] * phi[i];
                mu[i * primes[j]] = 0;
                d[i * primes[j]] = d[i] / (c[i] + 1) * (c[i] + 2);
                c[i * primes[j]] = c[i] + 1;
                break;
            }
            // phi[i * primes[j]] = (primes[j] - 1) * phi[i];
            mu[i * primes[j]] = - mu[i];
            d[i * primes[j]] = d[i] * 2;
            c[i * primes[j]] = 1;
        }
    }

    for(int i = 1; i <= n; i ++){
        sum[i] = sum[i - 1] + mu[i];
    }

    for(int i = 1; i <= n; i ++){
        sd[i] = sd[i - 1] + d[i];
    }
}


ll cal(int a, int b){
    ll res = 0;
    if(a > b) swap(a, b);
    for(int l = 1, r; l <= a; l = r + 1){
        r = min(a / (a / l), b / (b / l));
        res += (sum[r] - sum[l - 1]) * sd[a / l] * sd[b / l];
    }
    return res;
}


void solve()
{
    int n, m;
    sdd(n, m);
    printf("%lld\n", cal(n, m));

}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("/home/jungu/code/in.txt", "r", stdin);
    // freopen("/home/jungu/code/out.txt", "w", stdout);
    // freopen("/home/jungu/code/practice/out.txt","w",stdout);
#endif
    // ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
    sd(T);
    k = 50000;
    get(k);
    // init(10000000);
    // int cas = 1;
    while (T--)
    {
        // printf("Case #%d:", cas++);
        solve();
    }

    return 0;
}