P3455 [POI2007]ZAP-Queries(莫比乌斯反演)

题目

P3455 [POI2007]ZAP-Queries

解析

莫比乌斯反演。
给定\(n\)，\(m\)，\(d\)，求$$\sum_{i=1}^{{n}\sum_{j=1}}[gcd(i,j)=d]$$
那我们设$$f(x)=\sum_{i=1}^{{n}\sum_{j=1}}[gcd(i,j)=x]$$
设

\[\begin{aligned} F(x)=& \sum_{x\mid i}f(k) \\Q =&\sum_{x\mid k}\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=k]\\ &当gcd(i,j)=k且x\mid i时，会对答案做一次贡献;\\ &所以只要枚举gcd(i,j)，当x\mid (k=gcd(i,j))时，会对答案做一次贡献\\ =&\sum_{i=1}^{n}\sum_{j=1}^{m}[x\mid gcd(i,j)]\\ &\because x\mid gcd(i,j)\\ &\therefore x\mid i且x\mid j\\ =&\sum_{i=1}^{\frac{n}{x}}\sum_{j=1}^{\frac{m}{x}}[1\mid gcd(i,j)]\\ =&\lfloor \frac{n}{x}\rfloor \lfloor\frac{m}{x}\rfloor \end{aligned} \]

实在看不懂的话其实也可以这么理解，根据\(gcd\)的性质，发现\(F\)实际上就是在求有多少个\(ij\)都是\(x\)的倍数，\(1\)到\(n\)里有\(\lfloor\dfrac{n}{x}\rfloor\)个，\(1\)到\(m\)里有\(\lfloor\dfrac{m}{x}\rfloor\)个，根据乘法原理，就是\(\lfloor\dfrac{n}{x}\rfloor\lfloor\dfrac{m}{x}\rfloor\)。

然后直接反演

\[\begin{aligned} F(x)=& \sum_{x\mid i}f(i) \\ f(x)=&\sum_{x\mid i}\mu(\frac{i}{x})F(i)\\ =&\sum_{x\mid i}\mu(\frac{i}{x})\lfloor \frac{n}{i}\rfloor \lfloor\frac{m}{i}\rfloor \end{aligned} \]

将\(d\)带入

\[\begin{aligned} f(x)=\sum_{d\mid i}\mu(\frac{i}{d})\lfloor \frac{n}{i}\rfloor \lfloor\frac{m}{i}\rfloor \end{aligned} \]

令\(\dfrac{i}{d}=t\)，得到

\[\begin{aligned} f(d)=\sum_{t=1}^{min(a,b)}\mu(t)\lfloor \frac{n}{td}\rfloor \lfloor\frac{m}{td}\rfloor \end{aligned} \]

这样就做到了\(O(n)\)的做，最后套一个数论分块，就可以\(O(m\sqrt n)\)的做这道题

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int t, n, m, d, num, mx, ans;
int mu[N], p[N], sum[N];
bool vis[N];

template<class T>inline void read(T &x) {
    x = 0; int f = 0; char ch = getchar();
    while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    x = f ? -x : x;
    return;
}

void get_mu(int n) {
    mu[1] = 1;
    for (int i = 2; i <= n; ++i) {
        if (!vis[i]) p[++num] = i, mu[i] = -1;
        for (int j = 1; j <= num; ++j) {
            if (i * p[j] > n) break;
            vis[i * p[j]] = 1;
            if (i % p[j] == 0) {
                mu[i * p[j]] = 0;
                break;
            } else mu[i * p[j]] = -mu[i];
        }
    }
}

int main() {
    get_mu(N);
    for (int i = 1; i <= N; ++i) sum[i] = sum[i - 1] + mu[i];
    read(t);
    while (t--) {
        ans = 0;
        read(n), read(m), read(d);
        mx = min(n, m);
        for (int l = 1, r; l <= mx; l = r + 1) {
            r = min(n / (n / l), m / (m / l));
            ans += ((n / (l * d)) * (m / (l * d)) * (sum[r] - sum[l - 1]));
        }
        printf("%d\n", ans);
    }
    return 0;
}