bzoj 2820 / SPOJ PGCD 莫比乌斯反演

Posted 2020-10-06 ( m Lweleth)

那啥bzoj2818也是一样的,突然想起来好像拿来当周赛的练习题过，用欧拉函数写掉的。

求$(i,j)=prime$对数

\begin{eqnarray*}\sum_{i=1}^{n}\sum_{j=1}^{m}[(i,j)=p]&=&\sum_{p=2}^{min(n,m)}\sum_{i=1}^{\lfloor\frac{n}{p}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{p}\rfloor}[i⊥j]\newline&=&\sum_{p=2}^{min(n,m)}\sum_{i=1}^{\lfloor\frac{n}{p}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{p}\rfloor}\sum_{d|(i,j)}{\mu(d)}\newline&=&\sum_{p=2}^{min(n,m)}\sum_{d}^{\lfloor\frac{min(n,m)}{p}\rfloor}{\mu(d)}\lfloor\frac{n}{pd}\rfloor\lfloor\frac{m}{pd}\rfloor\end{eqnarray*}

枚举质数的倍数，预处理好，最后底数优化一下。

 

/** @Date    : 2017-09-09 00:24:45
  * @FileName: bzoj 2820 莫比乌斯反演.cpp
  * @Platform: Windows
  * @Author  : Lweleth ([email protected])
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e7+20;
const double eps = 1e-8;

int pri[N];
int mu[N];
LL sum[N];
int c = 0;
bool vis[N];

void mobius()
{
	MMF(vis);
	MMF(mu);
	mu[1] = 1;
	for(int i = 2; i < N; i++)
	{
		if(!vis[i])
			pri[c++] = i, mu[i] = -1;
		for(int j = 0; j < c && i * pri[j] < N; j++)
		{
			vis[i * pri[j]] = 1;
			if(i % pri[j] == 0)
			{
				mu[i * pri[j]] = 0;
				break;
			}
			else mu[i * pri[j]] = -mu[i];
		}
	}
	for(int i = 0; i < c; i++) //预处理 mu[dp/p]
		for(int j = 1; j * pri[i] < N; j++)
			sum[j * pri[i]] += mu[j];
	for(int i = 1; i < N; i++)
		sum[i] += sum[i - 1];
}

int main()
{
	mobius();
	int T;
	cin >> T;
	while(T--)
	{
		LL n, m;
		scanf("%lld%lld", &n, &m);
		LL ans = 0;
		LL mi = min(n, m);
		LL last;
		for(int i = 1; i <= mi; i = last + 1)
		{
			last = min(n/(n/i), m/(m/i));
			ans += (n / i) * (m / i) * (sum[last] - sum[i - 1]); 
		}
		printf("%lld\n", ans);
	}
    return 0;
}