2020百度之星初赛1 Function 莫比乌斯反演 (HDU 6750)

HDU6750 Function

题解

先将题意转化，我们得到
$$\sum _{i=1}^n\sum _{t|i}t[gcd(t,\frac{i}{t})=1]$$
化简这个式子即可

推导

$$\begin{array}{rl}\sum _{i=1}^n\sum _{t|i}t[gcd(t,\frac{i}{t})=1]& =\sum _{i=1}^n\sum _{t|i}t\sum _{d|t,dt|i}\mu (d)\\ & =\sum _{d=1}^n\mu (d)\sum _{t=1}^{\lfloor n/d\rfloor }td\sum _{i=1}^{\lfloor n/d^{2}t\rfloor }1\\ T=d^{2}t\to & =\sum _{T=1}^n\left(\sum _{i=1}^{\lfloor n/T\rfloor }1\right)\sum _{d^2|T}\mu (d)\frac{T}{d}\\ & =\sum _{T=1}^n\lfloor \frac{n}{T}\rfloor \sum _{d^2|T}\mu (d)\frac{T}{d}\\ & =\sum _{d=1}^{\sqrt{n}}\mu (d)d\sum _{T=1}^{\lfloor n/d^2\rfloor }\lfloor \frac{n/d^2}{T}\rfloor T\\ G(k)=\sum _{i=1}^k\lfloor \frac{k}{i}\rfloor i\to & =\sum _{d=1}^{\sqrt{n}}\mu (d)dG(\lfloor \frac{n}{d^2}\rfloor )\end{array}$$
这里G函数和原式子都可以用数论分块来做，总体复杂度$O(\sqrt{n}log(n))$

代码

#include 
using namespace std;
typedef long long ll;
const int MAX = 1e6 + 10;
const ll mod = 1e9 + 7;
const ll inv2 = 500000004;

int vis[MAX], prime[MAX], num, mu[MAX], f[MAX];
ll g[MAX];
void makeMobius(int siz) {
    mu[1] = 1, num = 0;
    for (int i = 2; i <= siz; i++) {
        if (!vis[i]) prime[++num] = i, mu[i] = -1;
        for (int j = 1; j <= num && i * prime[j] <= siz; j++) {
            vis[i * prime[j]] = 1;
            if (i % prime[j] == 0) {
                mu[i * prime[j]] = 0;
                break;
            }
            else mu[i * prime[j]] = mu[i] * mu[prime[j]];
        }
    }
    for (int d = 1; d <= siz; d++)
        for (int i = d; i <= siz; i += d)
            g[i] = (g[i] + d) % mod;
    for (int i = 1; i <= siz; i++) {
        f[i] = (f[i - 1] + 1ll * mu[i] * i % mod) % mod;//求mu * i的前缀和
        g[i] = (g[i] + g[i - 1]) % mod;
    }
}

ll sum(ll x) {
    x %= mod;
    return x * (x + 1) % mod * inv2 % mod;
}

ll calc(ll n) {
    if (n <= 1e6) return g[n];//没有提前筛这东西, T了一发
    ll res = 0;
    for (ll l = 1, r; l <= n; l = r + 1) {
        r = n / (n / l);
        res = (res + 1ll * (sum(r) - sum(l - 1) + mod) % mod * (n / l) % mod) % mod;
    }
    return res;
}

int main() {

    makeMobius(1e6);
    int T; scanf("%d", &T);
    while (T--) {
        ll n; scanf("%lld", &n);
        ll ans = 0;
        for (ll l = 1, r; l * l <= n; l = r + 1) {
            r = l;
            while (n / ((r + 1) * (r + 1)) == n / (r * r)) r++;//这里分块不太会，就这样写了
            ans = (ans + 1ll * calc(n / (l * l)) * ((f[r] - f[l - 1] + mod) % mod) % mod) % mod;
        }
        printf("%lld\n", (ans + mod) % mod);
    }

    return 0;
}