牛客-brz的函数-题解

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

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

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

[quadquadquadquadquadquad = quad sum_{k=1}^{n}{mu(k)} sum_{i=1}^{lfloor frac{n}{k} floor}{mu(ik)} sum_{j=1}^{lfloor frac{n}{k} floor}{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} floor)^{2})，得到下图(8 imes8)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 imes 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
", ans[n]);
    }
    return 0;
}