CCPC-Wannafly Winter Camp Day3 div2 F. 小清新数论* 莫比乌斯反演

小清新数论

心情：蒻蒻的第一道莫比乌斯反演！！看了好几个小时QAQ，终于看懂些了！开心！^_^
题解： $\sum_{i = 1}^n\sum_{j = 1}^n \mu(gcd(i,j)) \tag 1$
$\sum_{d = 1}^n\sum_{i = 1}^n\sum_{j = 1}^n\mu(d)[gcd(i,j) = =d]\tag 2$
$\sum_{d = 1}^n\mu(d)\sum_{i = 1}^n\sum_{j = 1}^n[gcd(i,j) = =d]\tag 3$
设 $g(d) =\sum_{i = 1}^n\sum_{j = 1}^n[gcd(i,j) = =d]$， $f(d)$为满足 $d|gcd(i,j)$的对数，那么我们就有
$f(d) = \sum_{i = 1}^{\lfloor \frac{n}{d}\rfloor}g(d\cdot i)\tag 4$
如果我们要求有多少 $gcd(i,j)$可以整除 $d$，那么必定 $i = d \cdot x, j = d \cdot x$，因此 $f(d) = \lfloor \frac{n}{d}\rfloor \cdot \lfloor \frac{n}{d} \rfloor\tag 5$
此时我们对 $(4)$进行莫比乌斯反演
$g(d) = \sum_{i = 1}^{\lfloor \frac{n}{d}\rfloor}f(d \cdot i)\mu(i)\tag 6$
$g(d)= \sum_{i = 1}^{\lfloor \frac{n}{d}\rfloor} {\lfloor \frac{n}{d\cdot i}\rfloor }^2\mu(i)\tag 7$
所以最后的 $ans =\sum_{d = 1}^{n}\mu(d)g(d)\tag 8$
最后再在求 $g(d)$和 $\mu(d)$的时候用除法分块就好了，总的复杂度为 $O(N)$。

代码

#include<bits/stdc++.h>
typedef long long LL;
using namespace std;
const int N = 1E7+10, mod = 998244353;
int prime[N], mu[N], cnt, n;
bool vis[N];
void get_M()
{
	mu[1] = 1;
	for(int i = 2; i < N; ++i) {
		if(vis[i] == 0) {
			prime[cnt++] = i;
			mu[i] = -1;
		}
		for(int j = 0; j < cnt && prime[j] * i < N; ++j) {
			vis[i * prime[j]] = 1;
			if(i % prime[j] == 0) break;
			mu[prime[j] * i] = -mu[i];
		}
	}
	for(int i = 1; i < N; ++i) {
		mu[i] = mu[i - 1] + mu[i];
	}
}

LL g(int d)
{
	LL ans = 0;
	for(int i = 1, last = 1; i <= n / d; i = last + 1) {
		last = n / (n / i);
		ans = (ans + (mu[last] - mu[i - 1] + mod) * (1LL * (n / d / i) * (n / d / i) % mod) ) % mod;
	}
	return ans;
}


int main()
{
#ifndef ONLINE_JUDGE
    freopen("input.in","r",stdin);
#endif
	get_M();
	LL ans = 0;
	cin >> n;
	for(int i = 1, last = 1; i <= n; i = last + 1) {
		last = n / (n / i);
		ans = (ans + g(i) * (mu[last] - mu[i - 1] + mod)) % mod;
	}
	cout << ans << endl;
    return 0;
}