P3327 [SDOI2015]约数个数和 莫比乌斯反演

链接

luogu

思路

第一个式子我也不会，luogu有个证明，自己感悟吧。
\[d(ij)=\sum\limits_{x|i}\sum\limits_{y|j}[gcd(x,y)==1]\]

\[\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}\sum\limits_{x|i}\sum\limits_{y|j}[gcd(x,y)==1]\]
\[\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}{\left \lfloor \frac{n}{i} \right \rfloor \left \lfloor \frac{m}{j} \right \rfloor \left [ gcd(i,j)==1 \right ]}\]
\[f(x)=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}{\left \lfloor \frac{n}{i} \right \rfloor \left \lfloor \frac{m}{j} \right \rfloor \left [ gcd(i,j)==x \right ]}\]
\[g(x)=\sum\limits_{x|d} f(d)\]
\[g(x)=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}{\left \lfloor \frac{n}{i} \right \rfloor \left \lfloor \frac{m}{j} \right \rfloor \left [ x|gcd(i,j)\right ]}\]
\[g(x)=\sum\limits_{i=1}^{\frac{n}{x}}\sum\limits_{j=1}^{\frac{m}{x}}{\left \lfloor \frac{n}{x*i} \right \rfloor \left \lfloor \frac{m}{x*j} \right \rfloor }\]
\[g(x)=\sum\limits_{i=1}^{\frac{n}{x}}{\left \lfloor \frac{n}{x*i} \right \rfloor }\sum\limits_{j=1}^{\frac{m}{x}}{\left \lfloor \frac{m}{x*j} \right \rfloor }\]

\[g(x)=\sum\limits_{i=1}^{N}{\left \lfloor \frac{N}{i} \right \rfloor }\sum\limits_{j=1}^{M}{\left \lfloor \frac{M}{j} \right \rfloor }(N=n/x,M=m/x)\]
整除分块预处理，O（1）查询g(x)
\[f(x)=\sum\limits_{x|d}\mu(\frac{d}{n})g(d)\]
所求\[f(1)=\sum\limits_{d=1}^{min(m,n)}\mu(d)g(d)\]
g是可以整除分块的

其他

改马蜂，加空格

代码

#include <bits/stdc++.h>
const int N = 5e5+7;
using namespace std;
int read() {
    int x = 0, f = 1; char s = getchar();
    for (;s > '9' || s < '0'; s = getchar()) if (s == '-') f = -1;
    for (;s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
    return x * f;
}
int n, m, T;
int pri[N], vis[N], tot, mu[N], g[N];
void Euler(int limit) {
    mu[1] = 1;
    for (int i = 2; i <= limit; ++i) {
        if (!vis[i]) {
            pri[++tot] = i;
            mu[i] = -1;
        }
        for (int j = 1; j <= tot && i * pri[j] <= limit; ++j) {
            vis[i * pri[j]] = 1;
            if (i % pri[j] == 0) {
                mu[i * pri[j]] = 0;
                break;
            }
            mu[i * pri[j]] = -mu[i];
        }
    }
    for (int i = 1; i <= limit; ++i) {
        for (int l = 1, r; l <= i; l = r + 1) {
            r = i / (i / l);
            g[i] += (r - l + 1) * (i / r);
        }
        mu[i] += mu[i - 1];
    }
}
void solve() {
    n = read(), m = read();
    if (n > m) swap(n, m);
    long long ans = 0;
    for (int l = 1, r; l <= n; l = r + 1) {
        r = min(n / (n / l), m / (m / l));
        ans += 1LL * (mu[r] - mu[l-1]) * (1LL * g[n/l] * g[m/l]);
    }
    printf("%lld\n", ans);
}
int main() {
    Euler(50000);
    int T = read();
    while (T--) solve();
    return 0;
}