「Luogu P2257」YY的GCD「莫比乌斯反演」

版权声明：本文为hzy原创文章，未经博主允许不可随意转载。 https://blog.csdn.net/Binary_Heap/article/details/82150286

题目传送门

题意

求 $\sum_{i=1}^{n} \sum_{j=1}^{m} [gcd(i, j)\;is\;prime]$

题解

不妨设$n \leq m$，推一下式子

$$\sum_{i=1}^{n} \sum_{j=1}^{m} [gcd(i, j)\;is\;prime]$$

$$=\sum_{p=1}^{n} \sum_{i=1}^{n} \sum_{j=1}^{m} [gcd(i, j) = p]$$

$$=\sum_{p=1}^{n} \sum_{i=1,p|i}^{n} \sum_{j=1,p|j}^{m} [gcd(\frac{i}{p}, \frac{j}{p}) = 1]$$

$$=\sum_{p=1}^{n} \sum_{i=1}^{\lfloor \frac{n}{p} \rfloor} \sum_{j=1}^{\lfloor \frac{m}{p} \rfloor} [gcd(i', j') = 1]$$

$$=\sum_{p=1}^{n} \sum_{i=1}^{\lfloor \frac{n}{p} \rfloor} \sum_{j=1}^{\lfloor \frac{m}{p} \rfloor} e(gcd(i', j'))$$

$$=\sum_{p=1}^{n} \sum_{i=1}^{\lfloor \frac{n}{p} \rfloor} \sum_{j=1}^{\lfloor \frac{m}{p} \rfloor} \sum_{d|gcd(i',j')} \mu(d)$$

$$=\sum_{p=1}^{n} \sum_{d=1}^{ \lfloor \frac{n}{p} \rfloor} \mu(d) \lfloor \frac{n}{pd} \rfloor \lfloor \frac{m}{pd} \rfloor$$

于是枚举$p$，线性筛求出莫比乌斯函数的前缀和，然后整除分块做

#include <cstring>
#include <cstdio>

typedef long long LL;

const int MAXN = 1e7 + 10;

bool tag[MAXN];
int pr[MAXN], tot, n, m;
int mu[MAXN], mus[MAXN];

void init(int n) {
    memset(tag, 1, sizeof tag);
    tag[1] = false;
    mu[1] = mus[1] = 1;
    for(int i = 2; i <= n; i ++) {
        if(tag[i]) {
            pr[++ tot] = i;
            mu[i] = -1;
        }
        for(int j = 1; j <= tot && pr[j] * (LL)i <= n; j ++) {
            tag[i * pr[j]] = false;
            if(i % pr[j] == 0) {
                mu[i * pr[j]] = 0;
                break ;
            }
            mu[i * pr[j]] = - mu[i];
        }
        mus[i] = mus[i - 1] + mu[i];
    }
}

LL calc(int r1, int r2) {
    LL ans = 0;
    int r = (r1 < r2 ? r1 : r2);
    for(int i = 1, j; i <= r; i = j + 1) {
        j = r1 / (r1 / i);
        if(j > r2 / (r2 / i)) j = r2 / (r2 / i);
        ans += (mus[j] - mus[i-1]) * 1ll * (r1 / i) * (r2 / i);
    }
    return ans;
}

int main() {
    init(1e7);
    int T;
    scanf("%d", &T);
    for(; T --; ) {
        scanf("%d%d", &n, &m);
        LL ans = 0;
        int x = (n < m ? n : m);
        for(int i = 1; i <= tot && pr[i] <= x; i ++)
            ans += calc(n / pr[i], m / pr[i]);
        printf("%lld\n", ans);
    }
    return 0;
}

于是：喜闻乐见$TLE$

考虑优化：

$$\sum_{p=1}^{n} \sum_{d=1}^{ \lfloor \frac{n}{p} \rfloor} \mu(d) \lfloor \frac{n}{pd} \rfloor \lfloor \frac{m}{pd} \rfloor$$

枚举$T=pd$：

$$=\sum_{T=1}^{n} \sum_{p|T}^{}\mu(\frac{T}{p}) \lfloor \frac{n}{T} \rfloor \lfloor \frac{m}{T} \rfloor$$

$$=\sum_{T=1}^{n} \lfloor \frac{n}{T} \rfloor \lfloor \frac{m}{T} \rfloor \sum_{p|T}^{}\mu(\frac{T}{p})$$

于是我们可以预处理后面的$\mu$，询问的时间复杂度从玄学的$O(?)$优化到$O(\sqrt{n} + \sqrt{m})$

#include <cstring>
#include <cstdio>

typedef long long LL;

const int MAXN = 1e7 + 10;

bool tag[MAXN];
int pr[MAXN], tot, n, m;
int mu[MAXN];
LL sum[MAXN];

void init(int n) {
    memset(tag, 1, sizeof tag);
    tag[1] = false;
    mu[1] = 1;
    for(int i = 2; i <= n; i ++) {
        if(tag[i]) {
            pr[++ tot] = i;
            mu[i] = -1;
        }
        for(int j = 1; j <= tot && pr[j] * (LL)i <= n; j ++) {
            tag[i * pr[j]] = false;
            if(i % pr[j] == 0) {
                mu[i * pr[j]] = 0;
                break ;
            }
            mu[i * pr[j]] = - mu[i];
        }
    }
    for(int i = 1; i <= tot; i ++)  
        for(int j = 1; j * 1ll * pr[i] <= n; j ++)
            sum[j * pr[i]] += mu[j];
    for(int i = 1; i <= n; i ++)
        sum[i] += sum[i - 1];
}

LL calc(int r1, int r2) {
    LL ans = 0;
    int r = (r1 < r2 ? r1 : r2);
    for(int i = 1, j; i <= r; i = j + 1) {
        j = r1 / (r1 / i);
        if(j > r2 / (r2 / i)) j = r2 / (r2 / i);
        ans += (sum[j] - sum[i-1]) * 1ll * (r1 / i) * (r2 / i);
    }
    return ans;
}

int main() {
    init(1e7);
    int T; scanf("%d", &T);
    for(; T --; ) {
        scanf("%d%d", &n, &m);
        if(n > m) n ^= m ^= n ^= m;
        printf("%lld\n", calc(n, m));
    }
    return 0;
}