牛客-brz的函数-题解

摘要：蒟蒻 \(\text{brz}\) 最近学习了莫比乌斯函数，由于学到了新东西蒟蒻 \(\text{brz}\) 感觉很高兴，一旁的神牛 \(\text{lzy}\) 十分不屑，随手丢了一道水题就难住了蒟蒻 \(\text{brz}\)...

对于原函数若 \(i,j\) 互质，则\(\mu(ij)=\mu(i)\mu(j)\)，否则\(\mu(ij)=0\)，于是有:

\[\sum_{i=1}^{n} \sum_{j=1}^{n}{\mu(ij)} \quad = \quad \sum_{i=1}^{n} \sum_{j=1}^{n}{\mu(i)\mu(j)} [gcd(i,j)=1] \]

\[\quad\quad\quad\quad\quad\quad = \quad \sum_{i=1}^{n} \sum_{j=1}^{n}{\mu(i)\mu(j)} \sum_{d|gcd(i,j)}{\mu(d)} \]

\[\quad\quad\quad\quad\quad\quad\quad\quad\quad = \quad \sum_{i=1}^{n} \sum_{j=1}^{n}{\mu(i)\mu(j)} \sum_{k=1}^{n}{\mu(k)} [k|gcd(i,j)] \]

\[\quad\quad\quad\quad\quad\quad = \quad \sum_{k=1}^{n}{\mu(k)} \sum_{i=1}^{\lfloor \frac{n}{k} \rfloor}{\mu(ik)} \sum_{j=1}^{\lfloor \frac{n}{k} \rfloor}{\mu(jk)} \]

令\(S(k,m) = \sum_{i = 1}^{m}{\mu(ik)}\)，代入得：\(\sum_{k=1}^{n} \mu(k) S(k,\lfloor \frac{n}{k} \rfloor)^{2}\)，得到下图\(8\times8\)的矩阵：

	1	2	3	4	5	6	7	8
1	1	1	1	1	1	1	1	1
2		1		1		1		1
3			1			1
4				1				1
5					1
6						1
7							1
8								1

图中值为1的位置为有效点，而在左上角\(i \times i\)内的有效点的值的累加和就是\(n=i\)时的原函数值。

S函数的值：\(S(k,m) = S(k,m-1) + \mu(mk)\)

有效点的值：\(val(k,i) = \mu(k)S(k,i)^{2} - \mu(k)S(k,i-1)^{2}\)

原函数的值：\(ans(n) = ans(n-1) + \sum_{d|n}{val(d,n)}\)

时间复杂度\(O(nln(n) + T)\)

#include<bits/stdc++.h>

using namespace std;

const int maxn = 5e4+1;
int prime[maxn], tot, mu[maxn];
bool vis[maxn];
vector<int> g[maxn];
int s[maxn], ans[maxn];

void seive(int n) {
    vis[1] = true, mu[1] = 1;
    for(int i = 2; i <= n; ++ i) {
        if (!vis[i]) prime[++tot] = i, mu[i] = -1;
        for(int j = 1; 1ll * i * prime[j] <= n; ++ j) {
            vis[i * prime[j]] = true;
            if (i % prime[j] == 0) break;
            mu[i*prime[j]] = mu[i] * mu[prime[j]];
        }
    }
}

int main() {
    seive(maxn-1);
    for(int i = 1; i <= maxn-1; ++ i) {
        for(int j = i; j <= maxn-1; j += i) {
            g[j].push_back(i);
        }
    }
    for(int i = 1; i <= maxn-1; ++ i) {
        ans[i] = ans[i-1];
        for(auto j: g[i]) {
            ans[i] -= mu[j] * s[j] * s[j];
            s[j] += mu[i];
            ans[i] += mu[j] * s[j] * s[j];
        }
    }
    int T; scanf("%d", &T);
    while(T --) {
        int n; scanf("%d", &n);
        printf("%d\n", ans[n]);
    }
    return 0;
}