洛谷P3327-约数个数和-莫比乌斯反演+约数个数函数前缀和

推式子

$$\begin{array}{r}ans=\sum _{i=1}^N\sum _{j=1}^{M}d(ij)\end{array}$$

现在有个式子：

$$\begin{array}{r}d(ij)=\sum _{x\mid i}\sum _{y\mid j}[\gcd (x,y)==1]\end{array}$$

所以：

$$\begin{array}{r}ans=\sum _{i=1}^N\sum _{j=1}^M\sum _{x\mid i}\sum _{y\mid j}[\gcd (x,y)==1]\end{array}$$

给$[\gcd (x,y)==1]$替换掉：

$$\begin{array}{r}ans=\sum _{d=1}^{\min (N,M)}\mu (d)\sum _{i=1}^N\sum _{j=1}^M\sum _{x\mid i}\sum _{y\mid j}[d\mid \gcd (x,y)]\end{array}$$

我们枚举$x$,$y$:

$$\begin{array}{r}ans=\sum _{d=1}^{\min (N,M)}\mu (d)\sum _{x=1}^N\sum _{y=1}^M[d\mid \gcd (x,y)]\lfloor \frac{N}{x}\rfloor \lfloor \frac{M}{y}\rfloor \end{array}$$

$$\begin{array}{r}ans=\sum _{d=1}^{\min (N,M)}\mu (d)\sum _{x=1}^{\lfloor \frac{N}{d}\rfloor }\sum _{y=1}^{\lfloor \frac{M}{d}\rfloor }\lfloor \frac{N}{xd}\rfloor \lfloor \frac{M}{yd}\rfloor \end{array}$$

$$\begin{array}{r}ans=\sum _{d=1}^{\min (N,M)}\mu (d)\sum _{x=1}^{\lfloor \frac{N}{d}\rfloor }\lfloor \frac{N}{xd}\rfloor \sum _{y=1}^{\lfloor \frac{M}{d}\rfloor }\lfloor \frac{M}{yd}\rfloor \end{array}$$

我们令$f(i)={\sum }_{i=1}^n\lfloor \frac{n}{i}\rfloor$

$$\begin{array}{r}ans=\sum _{d=1}^{\min (N,M)}\mu (d)f(\lfloor \frac{N}{i}\rfloor )f(\lfloor \frac{M}{i}\rfloor )\end{array}$$

$f(n)$就是约数个数函数的前缀和，我们预处理可以$O(1)$查询，然后分块求$ans$

#include 
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
template <typename T>
void out(T x) { cout << x << endl; }
ll fast_pow(ll a, ll b, ll p) {ll c = 1; while(b) { if(b & 1) c = c * a % p; a = a * a % p; b >>= 1;} return c;}
ll exgcd(ll a, ll b, ll &x, ll &y) { if(!b) {x = 1; y = 0; return a; } ll gcd = exgcd(b, a % b, y, x); y-= a / b * x; return gcd; }
const int N = 5e4 + 100;
int sum[N], d[N], mu[N], prime[N], tot;
//d[i]表示i的约数个数
//sum[i]表示i的最小质因子的最高次幂（线性筛用的是最小质因子去筛的）
bool mark[N];
void get_d_mu()
{
    mark[1] = mark[0] = true;
    tot = 0;
    mu[1] = d[1] = sum[1] = 1;
    for(int i = 2; i < N; i ++)
    {
        if(!mark[i])
        {
            prime[tot ++] = i;
            mu[i] = -1;
            d[i] = 2;
            sum[i] = 1;
        }
        for(int j = 0; j < tot && i * prime[j] < N; j ++)
        {
            mark[i * prime[j]] = true;
            if(i % prime[j] == 0)
            {
                d[i * prime[j]] = d[i] / (sum[i] + 1) * (sum[i] + 2);
                sum[i * prime[j]] = sum[i] + 1;
                mu[i * prime[j]] = 0;
                break;
            }
            mu[i * prime[j]] = -mu[i];
            d[i * prime[j]] = 2 * d[i];
            sum[i * prime[j]] = 1;
        }
    }
    for(int i = 1; i < N; i ++)
    {
        mu[i] += mu[i - 1];
        d[i] += d[i - 1];
    }
}
ll cal(ll n, ll m)
{
    if(n > m)
        swap(n, m);
    ll ans = 0;
    for(ll i = 1, j = 1; i <= n; i = j + 1)
    {
        j = min(n / (n / i), m / (m / i));
        ans += 1ll * (mu[j] - mu[i - 1]) * d[n / i] * d[m / i];
    }
    return ans;
}
int main()
{
    ios::sync_with_stdio(false);
    get_d_mu();
    int t;
    cin >> t;
    while(t --)
    {
        ll n, m;
        cin >> n >> m;
        cout << cal(n, m) << endl;
    }
}