题目描述
对于给出的 \(n\) 个询问，每次求有多少个数对 \((x,y)\)，满足 \(a \le x \le b\)，\(c \le y \le d\)，且 \(\gcd(x,y) = k\)，\(\gcd(x,y)\) 函数为 \(x\) 和 \(y\) 的最大公约数。
题解
莫比乌斯反演
我们可以用二维前缀和的思想，我们设
\(f(n,m)=\sum\limits_{i=1}^n \sum\limits_{j=1}^m [\gcd(i,j)=k]\)，
那答案应为
\(f(b,d)-f(b,c-1)-f(a-1,d)+f(a-1,c-1)\)
接下来看看\(f(n,m)\)怎么求：
\(\sum\limits_{i=1}^n \sum\limits_{j=1}^m [\gcd(i,j)=k]\)
\(=\sum\limits_{i=1}^{n/k} \sum\limits_{j=1}^{m/k} [\gcd(i,j)=1]\)
使用莫比乌斯反演
\(=\sum\limits_{i=1}^{n/k} \sum\limits_{j=1}^{m/k} \sum\limits_{d|gcd(i,j)}\mu(d)\)
把\(d\)放到前面枚举，设\(i=xd,\ j=yd\)
\(=\sum\limits_{d} \mu(d) * \sum\limits_{x=1}^{n/kd} \sum\limits_{y=1}^{m/kd} 1\)
\(=\sum\limits_{d} \mu(d) * \lfloor \frac{n}{kd}\rfloor * \lfloor \frac{m}{kd}\rfloor\)
预处理\(\mu(d)\)的前缀和，使用除法分块即可做到时间复杂度\(O(\sqrt{n})\)
总时间复杂度\(O(n\sqrt{n})\)
#include <bits/stdc++.h>
using namespace std;
int t, a, b, c, d, k;
int pr[50005], mb[50005], sum[50005], tot;
bool np[50005];
void init() {
mb[1] = 1;
for (int i = 2; i <= 50000; i++) {
if (!np[i]) pr[++tot] = i, mb[i] = -1;
for (int j = 1; j <= tot && i * pr[j] <= 50000; j++) {
np[i*pr[j]] = 1;
if (i % pr[j] == 0) {
mb[i*pr[j]] = 0;
break;
} else mb[i*pr[j]] = -mb[i];
}
}
for (int i = 1; i <= 50000; i++) sum[i] = sum[i-1] + mb[i];
}
int solve(int nn, int mm) {
int ret = 0, n = nn / k, m = mm / k;
for (int l = 1, r = 0; l <= min(n, m); l = r + 1) {
r = min(n / (n / l), m / (m / l));
ret += (sum[r] - sum[l-1]) * (n / l) * (m / l);
}
return ret;
}
int main() {
scanf("%d", &t);
init();
while (t--) {
scanf("%d %d %d %d %d", &a, &b, &c, &d, &k);
printf("%d\n", solve(b, d) + solve(a-1, c-1) - solve(b, c-1) - solve(a-1, d));
}
return 0;
}