【莫比乌斯反演】C - LCMs - AGC038

题目链接(https://atcoder.jp/contests/agc038/tasks/agc038_c)

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $700$ points

Problem Statement

We have an integer sequence of length $N$: $A_0,A_1,\cdots,A_{N-1}$.Find the following sum ($\mathrm{lcm}(a, b)$ denotes the least common multiple of $a$and $b$):$\sum_{i=0}^{N-2} \sum_{j=i+1}^{N-1} \mathrm{lcm}(A_i,A_j)$

Since the answer may be enormous, compute it modulo $998244353$.

Constraints

$1 \leq N \leq200000$

$1 \leq A_i \leq 1000000$

All values in input are integers.

Input

Input is given from Standard Input in the following format:

$N$

$A_0\ A_1\ \cdots\ A_{N-1}$

Output

Print the sum modulo $998244353$.

Sample Input 1

3 2 4 6

Sample Output 1

22

$\mathrm{lcm}(2,4)+\mathrm{lcm}(2,6)+\mathrm{lcm}(4,6)=4+6+12=22$.

Sample Input 2

8 1 2 3 4 6 8 12 12

Sample Output 2

313

Sample Input 3

10 356822 296174 484500 710640 518322 888250 259161 609120 592348 713644

Sample Output 3

353891724

题意

给定长度为$N$的序列$A_1,A_2,...,A_N$,求$\sum_{i = 0}^{N-2}\sum_{j=i+1}^{N-1}lcm(A_i.A_j)$，模$998244353$

$(1\leq N\leq 2*10^5,1\leq A_i\leq 10^6)$

思路

（笔记）

众所周知$lcm(a,b) == (a * b) / gcd(a,b)$

设答案为$ans$,则可以由$\sum_{i=0}^{N-1}\sum_{j=0}^{N-1}lcm(A_i,A_j) = 2 * ans + \sum_{i=0}^NA_i$

间接知道$ans$大小

$\sum_{i=0}^{N-1}\sum_{j=0}^{N-1}lcm(A_i,A_j)$

$=\sum_{i=0}^{N-1}\sum_{j=0}^{N-1}(A_i*A_j)/gcd(A_i,A_j)$

$=\sum_{d=1}^{L}1/d\sum_{i=0}^{N-1}\sum_{j=0}^{N-1}(A_i*A_j)[gcd(A_i,A_j) = d]$

设$f(d)=\sum_{i=0}^{N-1}\sum_{j=0}^{N-1}(A_i*A_j)[gcd(A_i,A_j) = d]$

对$f(d)$作倍数和,$i,j$相当于是一样的所以后面作平方

$g(d)=\sum_{d|d'}f(d')=(\sum_{i=0}^{N-1}A_i[d|A_i])^2$

$f(d)=\sum_{d|d'}μ(d'/d)g(d')=\sum_{d|d'}μ(d'/d)(\sum_{i=0}^{N-1}A_i[d|A_i])^2$

最终得到求和式

$\sum_{i=0}^{N-1}\sum_{j=0}^{N-1}lcm(A_i,A_j)=\sum_{d=1}^L1/d\sum_{d|d'}μ(d'/d)(\sum_{i=0}^{N-1}A_i[d'|A_i])^2$

在这里，对于任意的$d$的这样的一个求和式$h(d)=\sum_{d|n}f(d)g(n/d)$，可以通过枚举因子或者枚举倍数来求得$h(n)$,复杂度$O(nlogn)$

然后就可以一步步求求和式：

对于求和式，第一项是求所有的$A_i$是$d'$的倍数的求和， $(\sum_{i=0}^{N-1}A_i[d'|A_i])$，就可以把这里的求和看作$h(d') = \sum_{i = 0}^{N-1}A_i[d'|A_i]$,把$A_i$用一个数组存起来，对这个数组进行倍数和，可以用$nlogn$的复杂度求取$h(d')$，用桶数组$pos[]$优化方便累加，即$h(d') = \sum_{d'|i}pos_i$

然后就可以来求$\sum_{d|d'}μ(d'/d)h(d')^2$,又是类似于倍数和的一种形式，可以用$nlogn$的复杂度求取$\sum_{d|d'}μ(d'/d)$,即$h'(d) = \sum_{d|d'}μ(d'/d)$这样

最后可以用$O(n)$的复杂度求取$\sum_{d=1}^L1/d*h'(d)$

总复杂度$O(LlogL+N)，L=maxA_i$

（嘤嘤嘤，被笨死惹，不想敲注释，名字和思路里的变量都对应的上吧）

AC代码

#include <bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
#define llinf 0x3f3f3f3f3f3f3f3f
#define int long long
#define ull unsigned long long
#define PII pair<int,int>
#define endl '\n'
const int N = 1e6 + 100;
const int mod = 998244353;
const double pi = acos(-1.0);
typedef long long ll;
using namespace std;
int n;
int mx;
int a[N],pos[N];
int h1[N], h2[N];
int miu[N];
int prime[N], num;
bool vis[N];
int quickpow(int a, int b) {
	int res = 1;
	while (b > 0) {
		if (b & 1) res = res * a % mod;
		b >>= 1;
		a = a * a % mod;
	}
	return res;
}
int getinv(int a) {
	return quickpow(a, mod - 2);
}
void init() {
	mx = 0;
	for (int i = 1; i <= n; i++) {
		a[i] = 0;
	}
	return;
}
void solve() {
	init();
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
		pos[a[i]] = (pos[a[i]] + a[i]) % mod;
		mx = max(mx, a[i]);
	}

	miu[1] = 1;
	for (int i = 2; i <= 1e6; i++)
	{
		if (!vis[i]) prime[++num] = i, miu[i] = -1;
		for (int j = 1; j <= num && i * prime[j] <= 1e6; j++)
		{
			vis[i * prime[j]] = 1;
			if (i % prime[j] == 0) break;
			miu[i * prime[j]] = -miu[i];
		}
	}

	for (int d = 1; d <= 1e6; d++)
	{
		int d_inv = getinv(d);
		for (int dd = d; dd <= 1e6; dd += d)
		{
			h1[d] = (h1[d] + pos[dd]) % mod;
			h2[dd] = (h2[dd] + miu[dd / d] * d_inv % mod + mod) % mod;
		}
	}

	int ans = 0;
	for (int d = 1; d <= mx; d++) {
		ans = (ans + h1[d] * h1[d] % mod * h2[d] % mod) % mod;
	}

	ans = ((ans - h1[1]) % mod + mod) % mod;

	cout << ans * getinv(2) % mod;
	return;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	while (cin >> n) {
		solve();
	}
	return 0;
}