就是让你求(sumlimits_{i=1}sumlimits_{j=1}d(ij))(其中(d(x))表示(x)的因数个数)
首先有引理(然而并没有证明):
(d(ij)=sumlimits_{x|i}sumlimits_{y|j}[gcd(x,y)=1])
带到原式里得到:
(ans=sumlimits_{i=1}sumlimits_{j=1}sumlimits_{x|i}sumlimits_{y|j}[gcd(x,y)=1])
利用(mu)函数的性质把方括号换掉:
(ans=sumlimits_{i=1}sumlimits_{j=1}sumlimits_{x|i}sumlimits_{y|j}sumlimits_{d|gcd(x,y)}mu(d))
交换枚举主体:
(ans=sumlimits_{x=1}sumlimits_{y=1}sumlimits_{i=1}^{lfloorfrac{N}{x} floor}sumlimits_{j=1}^{lfloorfrac{M}{y} floor}sumlimits_{d|gcd(x,y)}mu(d))
进而得到:
(ans=sumlimits_{x=1}sumlimits_{y=1}lfloorfrac{N}{x} floorlfloorfrac{M}{y} floorsumlimits_{d|gcd(x,y)}mu(d))
首先枚举(d):
(ans=sumlimits_{d=1}^{min{N,M}}mu(d)sumlimits_{x=1}^{lfloorfrac{N}{d} floor}sumlimits_{y=1}^{lfloorfrac{M}{d} floor}lfloorfrac{N}{x} floorlfloorfrac{M}{y} floor)
后面的顺序是无所谓的,交换一下:
(ans=sumlimits_{d=1}^{min{N,M}}mu(d)sumlimits_{x=1}^{lfloorfrac{N}{d} floor}lfloorfrac{N}{x} floorsumlimits_{y=1}^{lfloorfrac{M}{d} floor}lfloorfrac{M}{y} floor)
然后发现只要预处理一下后面的东西就可以整除分块了
贴一下代码:
#include <bits/stdc++.h>
using namespace std;
#define N 50000
int cnt, prime[N+5], mu[N+5], sum[N+5], notprime[N+5];
int b[N+5];
void init()
{
mu[1] = sum[1] = notprime[1] = 1;
for(int i = 2; i <= N; ++i)
{
if(!notprime[i]) prime[++cnt] = i, mu[i] = -1;
for(int j = 1; j <= cnt && i*prime[j] <= N; ++j)
{
notprime[i*prime[j]] = 1;
if(i%prime[j] == 0)
{
mu[i*prime[j]] = 0;
break;
}
mu[i*prime[j]] = mu[i]*-1;
}
sum[i] = sum[i-1]+mu[i];
}
for(int i = 1; i <= N; ++i)
{
for(int l = 1, r; l <= i; l = r+1)
{
r = min(i/(i/l), i);
b[i] += (r-l+1)*(i/l);
}
}
}
int T, n, m;
int main()
{
scanf("%d", &T);
init();
while(T--)
{
scanf("%d%d", &n, &m);
long long ans = 0;
if(n > m) swap(n, m);
for(int l = 1, r; l <= n; l = r+1)
{
r = min(min(n/(n/l), m/(m/l)), n);
ans += 1LL*(sum[r]-sum[l-1])*b[n/l]*b[m/l];
}
printf("%lld
", ans);
}
return 0;
}