[SDOI2015]约数个数和

最新推荐文章于 2021-07-29 21:25:53 发布

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本文介绍了一种使用莫比乌斯反演解决特定数学问题的方法，即求解两个数乘积的约数个数之和。通过预处理μ函数前缀和与g(x)函数，实现了时间复杂度为O(T√x)的高效算法。

Description

$T$ 组询问。
设 $d (x)$ 为 $x$ 的约数个数，给定 $N$ 、 $M$ ，求 $∑i=1N∑j=1Md(ij)\sum_{i=1}^N \sum_{j=1}^M d(ij)$
$1≤n,m,T≤500001\le n, m, T\le 50000$

Solution

莫比乌斯反演即可：

$∑n=1N∑m=1Md(nm)\sum_{n = 1}^{N} \sum_{m = 1}^{M} d(nm)$

$=∑n=1N∑m=1M∑a∣n∑b∣m[gcd(a,b)==1]=\sum_{n = 1}^{N} \sum_{m = 1}^{M} \sum_{a|n} \sum_{b|m} [gcd(a,b) == 1]$

$=∑n=1N∑m=1M∑a∣n∑b∣m∑d∣gcd(a,b)μ(d)=\sum_{n = 1}^{N} \sum_{m = 1}^{M} \sum_{a|n} \sum_{b|m} \sum_{d|gcd(a,b)} \mu (d)$

$=∑d=1min⁡{N,M}μ(d)∑a=1⌊Nd⌋∑b=1⌊Md⌋⌊Nad⌋⌊Mbd⌋=\sum_{d = 1}^{\min\{N,M\}} \mu (d) \sum_{a = 1}^{\left \lfloor \frac{N}{d}\right \rfloor} \sum_{b = 1}^{\left \lfloor \frac{M}{d}\right \rfloor} \left \lfloor \frac{N}{ad}\right \rfloor \left \lfloor \frac{M}{bd}\right \rfloor$

$=∑d=1min⁡{N,M}μ(d)∑a=1⌊Nd⌋⌊Nad⌋∑b=1⌊Md⌋⌊Mbd⌋=\sum_{d = 1}^{\min\{N,M\}} \mu (d) \sum_{a = 1}^{\left \lfloor \frac{N}{d}\right \rfloor}\left \lfloor \frac{N}{ad}\right \rfloor \sum_{b = 1}^{\left \lfloor \frac{M}{d}\right \rfloor} \left \lfloor \frac{M}{bd}\right \rfloor$ 设 $\sum_{i=1}^x \left \lfloor \frac{x}{i} \right \rfloor$ 。

预处理 $g (x)$ ， $μ\mu$ 函数前缀和。

整除分块，时间复杂度 $O(Tx)O(T\sqrt x)$

Code

//Dlove's template
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#define R register
#define ll long long
#define ull unsigned long long
#define db double
#define ld long double
#define sqr(_x) ((_x) * (_x))
#define Cmax(_a, _b) ((_a) < (_b) ? (_a) = (_b), 1 : 0)
#define Cmin(_a, _b) ((_a) > (_b) ? (_a) = (_b), 1 : 0)
#define Max(_a, _b) ((_a) > (_b) ? (_a) : (_b))
#define Min(_a, _b) ((_a) < (_b) ? (_a) : (_b))
#define Abs(_x) (_x < 0 ? (-(_x)) : (_x))
using namespace std;
namespace Dntcry
{
	char Bs[1 << 22], *Ss = Bs, *Ts = Bs;		
	#define getchar() (Ss == Ts && (Ts = (Ss = Bs)  + fread(Bs, 1, 1 << 22, stdin), Ss == Ts)  EOF : *Ss++) 
	inline int read()
	{
		R int a = 0, b = 1; R char c = getchar();
		for(; c < '0' || c > '9'; c = getchar()) (c == '-') ? b = -1 : 0;
		for(; c >= '0' && c <= '9'; c = getchar()) a = (a << 1) + (a << 3) + c - '0';
		return a * b;
	}
	inline ll lread()
	{
		R ll a = 0, b = 1; R char c = getchar();
		for(; c < '0' || c > '9'; c = getchar()) (c == '-') ? b = -1 : 0;
		for(; c >= '0' && c <= '9'; c = getchar()) a = (a << 1) + (a << 3) + c - '0';
		return a * b;
	}
	const int Maxn = 50010;
	int T, n, m;
	int Pri[Maxn], tot, mu[Maxn];
	int g[Maxn];
	bool vis[Maxn];
	int getv(R int x)
	{
		R int res = 0;
		for(R int i = 1, next = 0; i <= x; i = next + 1)
		{
			next = x / (x / i);
			res += (next - i + 1) * (x / i);
		}
		return res;
	}
	int Main()
	{
		mu[1] = 1;
		for(R int i = 2; i <= 50000; i++)
		{
			if(!vis[i]) Pri[++tot] = i, mu[i] = -1;
			for(R int j = 1; j <= tot && Pri[j] * i <= 50000; j++)
			{
				vis[i * Pri[j]] = 1;
				if(i % Pri[j] == 0)
				{
					mu[i * Pri[j]] = 0;
					break;
				}
				else mu[i * Pri[j]] = -mu[i];
			}
			mu[i] += mu[i - 1];
		}
		for(R int i = 1; i <= 50000; i++) g[i] = getv(i);
		T = read();
		while(T--)
		{
			n = read(), m = read();
			if(n > m) swap(n, m);
			R ll Ans = 0;
			for(R int i = 1, next = 0; i <= n; i = next + 1)
			{
				next = Min(n / (n / i), m / (m / i));
				Ans += 1ll * (mu[next] - mu[i - 1]) * g[n / i] * g[m / i];
			}
			printf("%lld\n", Ans);
		}
		return 0;
	}
}
int main()
{
	return Dntcry :: Main();
}