[BZOJ3994] [SDOI2015] 约数个数和 [莫比乌斯反演][除法分块]

Link
https://www.lydsy.com/JudgeOnline/problem.php?id=3994

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Description
$\mathrm{T}$ 组数据，每一组数据给出 $n,m$ ，求 $\sum _{i=1}^n\sum _{j=1}^{m}d(ij)$ 。
$1\le n,m,\mathrm{T}\le 5\times 10^4$ 。

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Analysis
约数个数和显然可以变一下，得到
$\begin{array}{rcl}\sum _{i=1}^n\sum _{j=1}^{m}d(ij)& =& \sum _{i=1}^n\sum _{j=1}^m\prod _{k=1}^{\omega (ij)}(a_k-1)\end{array}$
但是这个没有什么用，我们考虑变求约数个数和为考虑每个数作为约数的贡献
为了方便下面我就当 $n=\min \{n,m\}$
$\begin{array}{rcl}\sum _{i=1}^n\sum _{j=1}^{m}d(ij)& =& \sum _{i=1}^n\sum _{j=1}^m[(i,j)=1]\lfloor \frac{n}{i}\rfloor \lfloor \frac{m}{j}\rfloor \\ & =& \sum _{i=1}^n\sum _{j=1}^mϵ((i,j))\lfloor \frac{n}{i}\rfloor \lfloor \frac{m}{j}\rfloor \\ & =& \sum _{i=1}^n\sum _{j=1}^m\sum _{d|(i,j)}\mu (d)\lfloor \frac{n}{i}\rfloor \lfloor \frac{m}{j}\rfloor \\ & =& \sum _{d=1}^n\mu (d)\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }ij\\ & =& \sum _{d=1}^n\mu (d)\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }\lfloor \frac{n}{di}\rfloor \lfloor \frac{m}{dj}\rfloor \\ & =& \sum _{d=1}^n\mu (d)\left(\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\lfloor \frac{\lfloor \frac{n}{d}\rfloor }{i}\rfloor \cdot \sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }\lfloor \frac{\lfloor \frac{m}{d}\rfloor }{j}\rfloor \right)\end{array}$
预处理 $f(x)=\sum _{i=1}^x\lfloor \frac{x}{i}\rfloor$ ，然后对每个询问再整除分块即可。$O(m\sqrt{m}+\mathrm{T}\sqrt{m})$

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这该死的莫反竟然如此的套路
oier平均一人一道约数个数和 noip出莫反裸题指日可待

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#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int MAXN = 5e4+5;
int T, n, m;
int mu[MAXN], smu[MAXN];
long long f[MAXN];
bool NotPrime[MAXN];
int PrimeList[MAXN];
int Diven[MAXN];
void Init()
{
	const int n = 5e4;
	for (register int i = 1; i <= n; ++i)
	{
		for (register int t, L = 1, R; L <= i; L = R + 1)
		{
			R = i / (t = i / L);
			f[i] += (R - L + 1) * t;
		}
	}
	NotPrime[0] = NotPrime[1] = 1;
	mu[1] = 1;
	for (register int i = 2; i <= n; ++i)
	{
		if (!NotPrime[i]) PrimeList[++PrimeList[0]] = i, Diven[PrimeList[0]] = n / i, mu[i] = -1;
		for (register int t, j = 1; j <= PrimeList[0] && i <= Diven[j]; ++j)
		{
			NotPrime[t = i * PrimeList[j]] = 1;
			if (i % PrimeList[j] == 0)
			{
				mu[t] = 0;
				break;
			}
			mu[t] = -mu[i];
		}
	}
	for (register int i = 1; i <= n; ++i)
	{
		smu[i] = smu[i-1] + mu[i];
	}
}
#define getchar() (frS==frT&&(frT=(frS=frBB)+fread(frBB,1,1<<12,stdin),frS==frT)?EOF:*frS++)
char frBB[1<<12], *frS=frBB, *frT=frBB;
inline int read(int& x)
{
	x=0;char ch=getchar();bool w=0;
	while(!isdigit(ch))w|=(ch=='-'),ch=getchar();
	while(isdigit(ch))x=x*10+(ch^48),ch=getchar();
	w?(x=-x):0;
}
int main()
{
	Init();
	read(T);
	register int fafa;
	register long long Ans;
	while (T--)
	{
		read(n); read(m);
		fafa = (n - m) >> 31;
		fafa = n & fafa | m & ~fafa;
		Ans = 0;
		for (register int tn, tm, L = 1, R; L <= fafa; L = R + 1)
		{
			R = min(n/(tn=n/L),m/(tm=m/L));
			Ans += f[tn] * f[tm] * (smu[R] - smu[L-1]);
		}
		printf("%lld\n", Ans);
	}
	return 0;
}