LibreOJ 2185.「SDOI2015」约数个数和

考虑把 d(xy)d(xy) 这个玩意拆开。
发现 d(xy)=ixjy[gcd(i,j)=1]d(xy) = \sum\limits_{i | x} \sum\limits_{j | y} [\gcd(i,j) = 1]
然鹅并不会证明……这里传送到 Siyuan 的证明
于是题目就是问的 i=1nj=1mxiyj[gcd(x,y)=1]\sum\limits_{i=1}^n \sum\limits_{j=1}^m \sum\limits_{x | i} \sum\limits_{y | j} [\gcd(x,y) = 1]

对于上面式子,我们可以转化一下思路:枚举 x,yx,y,计算其对给定范围内所有数的贡献。
于是有

然后设 f(x)=i=1nj=1mnimj[gcd(i,j)=x],F(x)=xif(i)=i=1nj=1mnimj[xgcd(i,j)]f(x) = \sum\limits_{i=1}^n \sum\limits_{j=1}^m \lfloor \dfrac n i \rfloor \lfloor \dfrac m j \rfloor [\gcd(i,j) = x],F(x) = \sum\limits_{x | i} f(i) = \sum\limits_{i=1}^n \sum\limits_{j=1}^m \lfloor \dfrac n i \rfloor \lfloor \dfrac m j \rfloor [x | \gcd(i,j)]

那么有 f(x)=xiμ(ix)F(i)f(x) = \sum\limits_{x | i} \mu(\dfrac i x) F(i),此题求 f(1)=i=1nμ(i)F(i)=i=1nμ(i)xinnxiyimmyif(1) = \sum\limits_{i=1}^n \mu(i) F(i) = \sum\limits_{i=1}^n \mu(i) \sum\limits_{xi \le n} \lfloor \dfrac {\lfloor \frac n x \rfloor} i \rfloor \sum\limits_{yi \le m} \lfloor \dfrac {\lfloor \frac m y \rfloor} i \rfloor

对于任意的 nn,预处理一下 i=1nni\sum\limits_{i = 1}^n \lfloor \dfrac n i \rfloorμ\mu 的前缀和然后数论分块。

代码:

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#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 5e4;
int t;
int cnt,prime[N + 5],mu[N + 5],vis[N + 5];
long long sum[N + 5];
int main()
{
mu[1] = 1;
for(register int i = 2;i <= N;++i)
{
if(!vis[i])
prime[++cnt] = i,mu[i] = -1;
for(register int j = 1;j <= cnt && i * prime[j] <= N;++j)
{
vis[i * prime[j]] = 1;
if(!(i % prime[j]))
break;
else
mu[i * prime[j]] = -mu[i];
}
}
for(register int i = 1;i <= N;++i)
{
mu[i] += mu[i - 1];
for(register int l = 1,r;l <= i;l = r + 1)
{
r = i / (i / l);
sum[i] += (long long)(i / l) * (r - l + 1);
}
}
scanf("%d",&t);
int n,m;
long long ans;
while(t--)
{
ans = 0;
scanf("%d%d",&n,&m);
if(n > m)
swap(n,m);
for(register int l = 1,r;l <= n;l = r + 1)
{
r = min(n / (n / l),m / (m / l));
ans += (long long)(mu[r] - mu[l - 1]) * sum[n / l] * sum[m / l];
}
printf("%lld\n",ans);
}
}