莫比乌斯反演
update in 2022.11
0.前言
老年退役选手的消遣
1.莫比乌斯函数
\(\mu\)或莫比乌斯函数是指以下函数：
\[\mu(n) = \left\{
\begin{aligned}
1 \quad\qquad\qquad \qquad \qquad \qquad \qquad n = 1
\\
\quad \qquad (-1)^k \qquad n = p_1 p_2...p_k, p_i为互不相同的素数
\\
0 \quad\qquad\qquad\qquad n含有指数大于1的质因子
\end{aligned}
\right.
\]
莫比乌斯函数的性质：
性质1：对于任意正整数n：
\[\epsilon(n) = \sum_{d\mid n} \mu(n) = \left\{
\begin{aligned}
1 \qquad n = 1
\\
0 \qquad n > 1
\end{aligned}
\right.
\]
性质2：对于任意正整数n：
\[\sum_{d \mid n} \frac{\mu(d)}{d} = \frac{\phi(n)}{n}
\]
证明：
即证
\[ n \cdot \sum_{d \mid n}\frac{\mu(d)}{d} = \phi(n)
\]
\[左 = n \cdot \sum_{d \mid n}\frac{\mu(d)}{d} = \sum_{d \mid n}\mu(d)\frac{n}{d} = \sum_{d \mid n}\mu(d)Id(\frac{n}{d}) = (\mu * Id)(n)\\
= (\mu * \phi * 1)(n) = (\mu * \phi * \mu^{-1})(n) = (\phi * \epsilon)(n) = \sum_{d \mid n}\phi(d)\epsilon(\frac{n}{d}) = \phi(n) = 右
\]
证毕
性质2推论：
\[\sum_{d \mid n} d\cdot \mu(\frac{n}{d}) = \phi(n)
\]
证明： 令\(k = \frac{n}{d}\)，带入原式，易得。
线性筛求 \(\mu\)：
lld mu[N];
void euler(lld n) {
lld p[N]; memset(p, 0, sizeof(p));
bool vis[N]; memset(vis, 0, sizeof(vis));
mu[1] = 1; int tot = 0;
for(int i = 2; i <= n; i++) {
if(!vis[i]) {
p[++tot] = i; vis[i] = true;
mu[i] = -1;
}
for(int j = 1; j <= tot && p[j] * i <= n; j++) {
vis[p[j] * i] = true;
if(i % p[j] == 0) mu[i * p[j]] = 0;
else mu[i * p[j]] = mu[i] * (-1);
}
}
}
2.莫比乌斯反演
设\(f(n), F(n)\) 为定义在正整数集合上的函数。
若 \(F(n)\) 满足：
\[F(n) = \sum_{d \mid n} f(d)
\]
则：
\[f(n) = \sum_{d \mid n} \mu(d)F(\frac{n}{d})
\]
或：
\[f(n) = \sum_{n \mid d} \mu(\frac{d}{n})F(d)
\]
前一个式子证明：（使用定义法）
\[右 = \sum_{d \mid n} \mu(d)F(\frac{n}{d}) = \sum_{d \mid n}\mu(d)\cdot \sum_{k \mid \frac{n}{d}}f(k) = \sum_{k \mid n} f(k) \cdot \sum_{d \mid \frac{n}{k}} \mu(d) = f(n) = 左
\]
证毕。（第三个等号看不明白可以代个数算算，可以发现两边正好相等）
也可以使用狄利克雷卷积证明
\[
F(n) = \sum_{d \mid n} f(d) = \sum_{d \mid n}f(d) \cdot 1(\frac{n}{d}) = (f * 1)(n) = (f * \mu^{-1})(n)
\]
\[
f(n) = \sum_{d \mid n}f(d) \cdot \epsilon(\frac{n}{d}) = (f * (\mu * \mu^{-1}))(n) = (\mu * (f * \mu^{-1}))(n) \\
= \sum_{d \mid n} \mu(d)(f * \mu^{-1})(\frac{n}{d}) = \sum_{d \mid n} \mu(d) F(\frac{n}{d})
\]
证毕
3.题目：
求：
\[\sum_{i = a}^{b} \sum_{j = c}^{d}[gcd(i, j) = k]
\]
利用二维前缀和将其转化为求
\[\sum_{i = 1}^{n} \sum_{j = 1}^{m}[gcd(i, j) = k]
\]
开始推式子：(式子中的除法均为整除）
\[
\begin{aligned}
&& \sum_{i = 1}^{n} \sum_{j = 1}^{m}[gcd(i, j) = k]
&& = \sum_{i = 1}^{ \frac{n}{k} } \sum_{j = 1}^{ \frac{m}{k} }[gcd(i, j) = 1]
&& = \sum_{i = 1}^{ \frac{n}{k}} \sum_{j = 1}^{ \frac{m}{k}} \epsilon(gcd(i, j))
&& = \sum_{i = 1}^{ \frac{n}{k} } \sum_{j = 1}^{ \frac{m}{k} } \sum_{d \mid gcd(i, j)} \mu(d)
\end{aligned}
\]
因为 \(d \mid gcd(i, j)\) 可以表示为同时满足 \(d \mid i, d \mid j\) ，所以
\[\begin{aligned}
&& 原式 = \sum_{d = 1}^{min(\frac{n}{k}, \frac{m}{k})}\mu(d) \cdot \sum_{d \mid i}^{\frac{n}{k}}\sum_{d \mid j}^{\frac{m}{k}} 1
&& = \sum_{d = 1}^{min(\frac{n}{k}, \frac{m}{k})}\mu(d) \cdot \sum_{i = 1}^{\frac{n}{kd}}\sum_{j = 1}^{\frac{m}{kd}} 1
&& = \sum_{d = 1}^{min(\frac{n}{k}, \frac{m}{k})}\mu(d) \frac{n}{kd}\frac{m}{kd}
\end{aligned}
\]
至此，已经可以 \(O(n)\) 求解了。但题目中询问次数很大，所以需要降低每次询问的复杂度。
考虑整除分块，设当前一段区间的左端点为 \(d\) ，则右端点为 \(min(n/(n / d), m / (m / d))\), \(\mu(d)\) 用前缀和实现。
typedef int lld;
const int N = 100005;
lld mu[N];
void euler(lld n) {
lld p[N]; memset(p, 0, sizeof(p));
bool vis[N]; memset(vis, 0, sizeof(vis));
mu[1] = 1; int tot = 0;
for(int i = 2; i <= n; i++) {
if(!vis[i]) {
p[++tot] = i; vis[i] = true;
mu[i] = -1;
}
for(int j = 1; j <= tot && p[j] * i <= n; j++) {
vis[p[j] * i] = true;
if(i % p[j] == 0) mu[i * p[j]] = 0;
else mu[i * p[j]] = mu[i] * (-1);
}
}
for(int i = 1; i <= n; i++) {
mu[i] += mu[i - 1];
}
}
lld a[N], b[N], c[N], d[N], k[N];
lld t, f[N];
lld calc(lld n, lld m, lld k) {
lld p = min(n / k, m / k), tmp = 0;
for(lld i = 1; i <= p; ) {
lld j = min(n / (n / i), m / (m / i));
if(j > p) break;
tmp = tmp + (mu[j] - mu[i - 1]) * (n / i / k) * (m / i / k);
i = j + 1;
}
return tmp;
}
int main() {
// freopen("data.in", "r", stdin);
int n; scanf("%d", &n);
for(int i = 1; i <= n; i++) {
scanf("%d%d%d%d%d", &a[i], &b[i], &c[i], &d[i], &k[i]);
t = max(t, max(b[i], d[i]));
}
euler(t);
for(int i = 1; i <= n; i++) {
printf("%d\n", calc(b[i], d[i], k[i]) - calc(b[i], c[i] - 1, k[i]) - calc(a[i] - 1, d[i], k[i]) + calc(a[i] - 1, c[i] - 1, k[i]));
}
return 0;
}
此题开long long会暴毙。
