洛谷P3327 约数个数和 结论+莫比乌斯反演

原题

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;
}