牛客挑战赛47 F-简单题 莫比乌斯反演

传送门： https://ac.nowcoder.com/acm/contest/10743/F

题意

$$求解\sum _{i=1}^n\sum _{j=1}^{m}gcd(i,j)\mu (ij)\phi (ij)$$

思路

$根据积性函数和\mu 的性质，只有i和j互质时才有贡献，不然\mu (ij)=0$
$i和j互质时，\phi (ij)=\phi (i)\phi (j)\mu (ij)=\mu (i)\mu (j)gcd(i,j)=1$

$根据上述可得：$

$$\sum _{i=1}^n\sum _{j=1}^m[gcd(i,j)=1]\phi (i)\mu (i)\phi (j)\mu (j)$$
$设f(i)=\phi (i)\ast \mu (i)，g(j)=\phi (j)\ast \mu (j)，则$

$$\sum _{i=1}^n\sum _{j=1}^m[gcd(i,j)=1]f(i)g(j)$$

$$\sum _{i=1}^n\sum _{j=1}^{m}f(i)g(j)\sum _{d|id|j}\mu (d)$$

$交换枚举顺序：$

$$\sum _{d=1}^{min(n,m)}\mu (d)\sum _{i=1}^{n}f(i)[d|i]\sum _{j=1}^{m}g(j)[d|j]$$

$根据上式，可以预处理\phi ,\mu ,f,g函数。然后暴力求解，即：$

		int n, m; cin >> n >> m;
        for(int i = 1;i <= n; i++) f[i] = phi[i];
        for(int j = 1;j <= m; j++) g[j] = phi[j];
        ll ans = 0;

        for(int d = 1;d <= min(n, m); d++) {
            if(!mu[d]) continue;
            ll sum1 = 0, sum2 = 0;
            for(int i = d;i <= n; i += d) sum1 += f[i];
            for(int j = d;j <= m; j += d) sum2 += g[j];
            ans = (ans + sum1 * sum2 * mu[d]) ;
        }

$但是这样会T掉，因为d=1时，n过大，那复杂度就是O(n^2)了。$

$需要Dirichlet前缀和优化即可。这是前缀和的几个版本：$Dirichlet 前缀和

$Dirichlet前缀和优化之后的复杂度为O(n\log \log n)，完全可以接受。$

Code(2512 MS)

#include 

using namespace std;

typedef long long ll;

const ll mod = (1ll << 32);
const int N = 2e7 + 10;

int cnt;
int prime[N];
bool is_prime[N];
int mu[N];
int phi[N];

void init() {
    phi[1] = mu[1] = is_prime[0] = is_prime[1] = 1;
    for(int i = 2;i < N; i++) {
        if(!is_prime[i]) {
            prime[++cnt] = i;
            phi[i] = i - 1;
            mu[i] = -1;
        }
        for(int j = 1;j <= cnt && prime[j] * i < N; j++) {
            is_prime[i * prime[j]] = 1;
            if(i % prime[j] == 0) {
                mu[i * prime[j]] = 0;
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
            phi[i * prime[j]] = phi[i] * (prime[j] - 1);
            mu[i * prime[j]] = -mu[i];
        }
    }

    for(int i = 1;i < N; i++) phi[i] *= mu[i];

}

int f[N], g[N];

void solve() {
    int _; cin >> _;
    init();
    while(_--) {
        int n, m; cin >> n >> m;
        for(int i = 1;i <= n; i++) f[i] = phi[i];
        for(int j = 1;j <= m; j++) g[j] = phi[j];
        
        ll ans = 0;
        
        for(int i = 1;i <= cnt; i++) {
            for(int j = n / prime[i];j >= 1; j--) f[j] += f[j * prime[i]];
            for(int j = m / prime[i];j >= 1; j--) g[j] += g[j * prime[i]];
        }

        for(int i = 1;i <= min(n, m); i++) {
            ans += 1ll * f[i] * g[i] * mu[i] % mod;
            ans %= mod;
        }
        cout << (ans + mod) % mod << endl;
    }
}

signed main() {
    solve();
    return 0;
}