P3327 [SDOI2015]约数个数和 (mobius反演)

P3327 [SDOI2015]约数个数和

推导过程

求 $\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} d(ij)$

$= \sum_{i = 1} ^{n} \sum_{j = 1} ^{m} \sum_{x \mid i} \sum_{y \mid j} gcd(x, y) == 1$

改成枚举 $x, y$，

$= \sum_{i = 1} ^{n} \sum_{j = 1} ^{m} \sum _{i \mid x} \sum_{j \mid y} gcd(i, j) == 1$

$= \sum_{i = 1} ^{n} \sum_{j = 1} ^{m} \lfloor\frac{n}{i}\rfloor \lfloor\frac{m}{j}\rfloor (gcd(i, j) == 1)$

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

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

最后我们只要预处理一下所有 $\frac{n}{d} \frac{m}{d}$的整除分块即可，整体复杂度 $n \sqrt n$

代码

/*
  Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>

#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;

inline ll read() {
    ll f = 1, x = 0;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-')    f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f * x;
}

const int N = 5e4 + 10;

bool st[N];

vector<int> prime;

int n, m, mu[N];

ll sum[N];

void mobius() {
    st[0] = st[1] = mu[1] = 1;
    for(int i = 2; i < N; i++) {
        if(!st[i]) {
            prime.pb(i);
            mu[i] = -1;
        }
        for(int j = 0; j < prime.size() && i * prime[j] < N; j++) {
            st[i * prime[j]] = 1;
            if(i % prime[j] == 0) break;
            mu[i * prime[j]] = -mu[i];
        }
    }
    for(int i = 1; i < N; i++) {
        mu[i] += mu[i - 1];
        for(ll l = 1, r; l <= i; l = r + 1) {
            r = i / (i / l);
            sum[i] += (i / l) * (r - l + 1);
        }
    }
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    mobius();
    int T = read();
    while(T--) {
        ll n = read(), m = read(), ans = 0;
        if(n > m) swap(n, m);
        for(ll l = 1, r; l <= n; l = r + 1) {
            r = min(n / (n / l), m / (m / l));
            ans += 1ll * (mu[r] - mu[l - 1]) * sum[n / l] * sum[m / l];
        }
        printf("%lld\n", ans);
    }
	return 0;
}