「BZOJ 4176」Lucas的数论-莫比乌斯反演+杜教筛


$$\sum_{i = 1} ^ n\sum_{j = 1} ^ nd(ij)$$

链接

BZOJ 4176

题解

首先有一个结论
$$d(nm) = \sum_{i | n}\sum_{j | m}[\gcd(i, j) = 1]$$
证明:设 $nm = p_1^{x_1}\cdot p_2^{x_2}\cdot p_3^{x_3}\cdots p_k^{x_k}$,$n = p_1^{y_1}\cdot p_2^{y_2}\cdot p_3^{y_3}\cdots p_k^{y_k}$,则
$m = p_1^{x_1 - y_1}\cdot p_2^{x_2 - y_2}\cdot p_3^{x_3 - y_3}\cdots p_k^{x_k - y_k}$。
设 $i = p_1^{a_1}\cdot p_2^{a_2}\cdot p_3^{a_3}\cdots p_k^{a_k}$,$j = p_1^{b_1}\cdot p_2^{b_2}\cdot p_3^{b_3}\cdots p_k^{b_k}$,我们先只考虑 $p_1$(其余可以同理),要使 $\gcd(i, j) = 1$,必然有 $a_1 = 0$ 或 $b_1 = 0$,若 $a_1 = 0$,那么 $b_1$ 有 $x_1 - y_1 + 1$ 种取值;如果 $b_1 = 0$,那么 $a_1$ 有 $y_1 + 1$ 种取值,一共 $x_1 + 1$ 种取值,对其他 $x_i$ 也是如此。
所以满足条件的 $i, j$ 对数为 $\prod(x_i + 1)$,即与约数定理的形式一样。

故原式可以变为
$$\begin{aligned}\sum_{i = 1} ^ n\sum_{j = 1} ^ n\sum_{x | i}\sum_{y | j}[\gcd(x, y) = 1] &= \sum_{i = 1} ^ n\sum_{j = 1} ^ n\sum_{x | i}\sum_{y | j}\sum_{d | x, d | y}\mu(d) \\ &= \sum_{d=1}^n\mu(d)\sum_{x=1}^n[d|x]\sum_{i=1}^n[x|i]\sum\limits_{y=1}^n[d|y]\sum_{j=1}^n[y|j] \\ &= \sum_{d=1}^n\mu(d)(\sum_{x=1}^{\lfloor \frac n d\rfloor}\sum_{i=1}^{\lfloor\frac n d\rfloor}[x|i])^2 \\ &= \sum_{d = 1} ^ n\mu(d)(\sum_{x = 1} ^ {\lfloor \frac n d \rfloor}\lfloor \frac {\lfloor \frac n d \rfloor} {x} \rfloor) ^ 2 \end{aligned}$$

然后 $\mu(d)$ 可以用杜教筛计算,后面的可以分块套分块计算,时间复杂度为 $O(n ^ {\frac 2 3} + n ^ {\frac 3 4})$

代码

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/**
* Copyright (c) 2017, xehoth
* All rights reserved.
* 「BZOJ 4176」Lucas的数论 18-08-2017
* 杜教筛
* @author xehoth
*/
#include <bits/stdc++.h>

namespace Task {

#define long long long

const int MAX_SIEVE_N = 1000000;
const int MOD = 1000000007;

int mu[MAX_SIEVE_N + 1], prime[MAX_SIEVE_N], cnt;
bool vis[MAX_SIEVE_N + 1];

inline void fastLinearSieveMethod(const int n) {
mu[1] = 1;
for (register int i = 2; i <= n; i++) {
if (!vis[i]) prime[cnt++] = i, mu[i] = -1;
for (register int j = 0, t; j < n && (t = i * prime[j]) <= n; j++) {
vis[t] = true;
if (i % prime[j] == 0) {
mu[t] = 0;
break;
} else {
mu[t] = -mu[i];
}
}
}
for (register int i = 2; i <= n; i++) mu[i] += mu[i - 1];
}

const int BLOCK_SIZE = 31623, BOUND = 1e9;
int buc1[BLOCK_SIZE + 1], buc2[BLOCK_SIZE + 1];

inline int &get(int x) { return x < BLOCK_SIZE ? buc1[x] : buc2[BOUND / x]; }

inline int sieve(int x) {
if (x <= MAX_SIEVE_N) return mu[x];
register int &cur = get(x);
if (cur) return cur;
register int ret = 1;
for (register int i = 2, pos; i <= x; i = pos + 1) {
pos = x / (x / i);
ret = (ret - (pos - i + 1) * (long)sieve(x / i) % MOD + MOD) % MOD;
}
return cur = ret;
}

inline int getF(int x) {
register int ret = 0;
for (register int i = 1, pos; i <= x; i = pos + 1) {
pos = x / (x / i);
ret = (ret - (pos - i + 1ll) * (x / i) % MOD + MOD) % MOD;
}
return (long)ret * ret % MOD;
}

inline void solve() {
std::ios::sync_with_stdio(false), std::cin.tie(NULL), std::cout.tie(NULL);
register int n, ans = 0;
std::cin >> n;
fastLinearSieveMethod(MAX_SIEVE_N);
for (register int i = 1, pos; i <= n; i = pos + 1) {
pos = n / (n / i),
ans = (ans +
(sieve(pos) - sieve(i - 1) + (long)MOD) % MOD * getF(n / i) %
MOD) %
MOD;
}
std::cout << ans;
}

#undef long
}

int main() {
Task::solve();
return 0;
}
# Math

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