「POI2007」ZAP-Queries「莫比乌斯反演」

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#莫比乌斯反演

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博客围绕莫比乌斯反演相关题目展开，包含T组询问求解问题。题解中假设a≤b，采用整除分块的方法来解决题目。

题目传送门

题意

T组询问求，

\sum i = 1 a \sum j = 1 b [g c d (i, j) = d]

$\sum_{i=1}^{a} \sum_{j=1}^{b} [gcd(i, j)=d]$ ，

T,a,b,d≤5×104 T , a , b , d ≤ 5 × 10 4 $T,a,b,d\leq 5 \times 10^4$

题解

不妨设 $a \leq b$ .

\sum i = 1 a \sum j = 1 b [g c d (i, j) = d]

$\sum_{i=1}^{a} \sum_{j=1}^{b} [gcd(i, j)=d]$

= \sum i = 1 ⌊ a d ⌋ \sum j = 1 ⌊ b d ⌋ [g c d (i', j') = 1]

$=\sum_{i=1}^{\lfloor\frac{a}{d}\rfloor} \sum_{j=1}^{\lfloor\frac{b}{d}\rfloor} [gcd(i', j')=1]$

= \sum i = 1 ⌊ a d ⌋ \sum j = 1 ⌊ b d ⌋ e (g c d (i, j))

$=\sum_{i=1}^{\lfloor\frac{a}{d}\rfloor} \sum_{j=1}^{\lfloor\frac{b}{d}\rfloor} e(gcd(i, j))$

= \sum i = 1 ⌊ a d ⌋ \sum j = 1 ⌊ b d ⌋ \sum k | g c d (i, j) e (g c d (i, j))

$=\sum_{i=1}^{\lfloor\frac{a}{d}\rfloor} \sum_{j=1}^{\lfloor\frac{b}{d}\rfloor}\sum_{k|gcd(i,j)}e(gcd(i, j))$

= \sum i = 1 ⌊ a d ⌋ \sum j = 1 ⌊ b d ⌋ \sum k | i, k | j μ (k)

$=\sum_{i=1}^{\lfloor\frac{a}{d}\rfloor} \sum_{j=1}^{\lfloor\frac{b}{d}\rfloor}\sum_{k|i,k|j}\mu(k)$

= \sum k = 1 ⌊ a d ⌋ μ (k) ⌊ a d k ⌋ ⌊ b d k ⌋

$=\sum_{k=1}^{\lfloor\frac{a}{d}\rfloor}\mu(k) \lfloor\frac{a}{dk}\rfloor \lfloor\frac{b}{dk}\rfloor$

然后整除分块做一下就行.

#include <iostream>
#include <cstdio>
using namespace std;

typedef long long LL;

const int MAXN = 50010;

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

void init(int n) {
    tag[1] = true;
    mus[1] = 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 && i * pr[j] <= n; j ++) {
            tag[i * pr[j]] = true;
            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 a, int b, int d) {
    if(a > b) a ^= b ^= a ^= b;
    a /= d; b /= d;
    LL ans = 0;
    for(int i = 1, j; i <= a; i = j + 1) {
        j = min(a / (a / i), b / (b / i));
        ans += (mus[j] - mus[i - 1]) * (a / i) * (b / i);
    }
    return ans;
}

int main() {
    init(50000);
    int n, a, b, d;
    for(scanf("%d", &n); n --; ) {
        scanf("%d%d%d", &a, &b, &d);
        printf("%lld\n", calc(a, b, d));
    }
    return 0;
}