Meaning of the questions:
Find x in [1, n] range, y in [1, m] in the range satisfying gcd (x, y) = k x, y, logarithmic.
answer:
-
If f (k) within a range of topics logarithmic gcd (x, y) = k, but this f (k) Comparison of "hard to find." We found that gcd (x, y)% k == x 0 , y, but a good number of requirements. Let's express gcd (x, y)% k == 0 with a number of F (k). In this case I wanted to be able to seek f (k) by F (k). After \ (F (k) = \ sum_ {k | d} f (d) \) which can be used to give Mobius inversion formula
\[f(k) = \sum_{k|d} \mu(\frac{d}{k})F(d) \]\[f(k) = \sum_{k|d} \mu(\frac{d}{k}) \lfloor\frac{n}{d}\rfloor \lfloor\frac{m}{d}\rfloor \]We iterate over all d (d is a multiple of k) to sue for peace. Because if \ (GCD (X, Y) =. 1 \) , then \ (gcd (k * x, k * y) = k \) so we can find the above formula f (1)
\[f(1) = \sum_{d=1}^{lim} \mu(d) \lfloor\frac{n}{d}\rfloor \lfloor\frac{m}{d}\rfloor \]Because k * x <= n, k * y <= m. Therefore lim = min (n / k, m / k)
Complexity of O (n).
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
const int N = 1e6 + 5;
int T, k, a, b, c, d;
int pri[N], cnt;
int mu[N];
int vis[N];
void prime(){
mu[1] = 1;
for(int i = 2; i <= 4e5; ++ i){
if(!vis[i]){
pri[++ cnt] = i;
mu[i] = -1;
}
for(int j = 1; j <= cnt && pri[j] * i <= 4e5; ++ j){
vis[pri[j] * i] = 1;
if(i % pri[j] == 0){
mu[pri[j] * i] = 0;
break;
}
mu[pri[j] * i] = -mu[i];
}
}
}
int main()
{
prime();
scanf("%d",&T);
for(int Case = 1; Case <= T; ++ Case){
scanf("%d%d%d", &b, &d, &k);
if(!k){
printf("0\n", Case);
continue;
}
b /= k, d /= k;
if(b > d) swap(b, d);
ll ans = 0, ans1 = 0;
for(int i = 1; i <= b; ++ i) ans += (ll)mu[i] * (b / i) * (d / i);
for(int i = 1; i <= b; ++ i) ans1 += (ll)mu[i] * (b / i) * (b / i);
ans -= ans1 / 2;
printf("%lld\n", ans);
}
return 0;
}
Problem solution blog: https://blog.csdn.net/litble/article/details/72804050?depth_1-utm_source=distribute.pc_relevant.none-task&utm_source=distribute.pc_relevant.none-task