算术 HDU

大意: 给定$n,m$, 求$sumlimits_{i=1}^nsumlimits_{j=1}^mmu(lcm(i,j))$

首先有$mu(lcm(i,j))=mu(i)mu(j)mu(gcd(i,j))$

枚举$gcd$可以得到$sumlimits_{d=1}^{min(n,m)}mu(d)sumlimits_{i=1}^{lfloorfrac{n}{d} floor}sumlimits_{j=1}^{lfloorfrac{m}{d} floor}mu(id)mu(jd)[gcd(i,j)=1]$

再反演一下就有$sumlimits_{d_1=1}^{min(n,m)}mu(d_1)sumlimits_{d_2=1}^{lfloorfrac{min(n,m)}{d_1} floor}mu(d_2)sumlimits_{i=1}^{lfloorfrac{n}{d_1d_2} floor}sumlimits_{j=1}^{lfloorfrac{m}{d_1d_2} floor}mu(id_1d_2)mu(jd_1d_2)$

枚举$d_1d_2$就有$sumlimits_{T=1}^{min(n,m)}sumlimits_{dmid T}mu(d)mu(frac{T}{d})sumlimits_{i=1}^{lfloorfrac{n}{T} floor}mu(iT)sumlimits_{j=1}^{lfloorfrac{m}{T} floor}mu(jT)$

预处理一下$mu * mu$即可

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 1e6+50;
int cnt, p[N], mu[N], mu2[N], vis[N];

void init() {
	mu[1] = mu2[1] = 1;
	REP(i,2,N-1) {
		if (!vis[i]) p[++cnt]=i,mu[i]=-1,mu2[i]=-2;
		for (int j=1,t; j<=cnt&&i*p[j]<N; ++j) {
			vis[t=i*p[j]] = 1;
			if (i%p[j]==0) {
				mu[t] = 0;
				if (i/p[j]%p[j]==0) mu2[t] = 0;
				else mu2[t] = 1;
				break;
			}
			mu[t] = -mu[i];
			mu2[t] = -2*mu2[i];
		}
	}
}

void work() {
	int n, m;
	scanf("%d%d", &n, &m);
	int mx = min(n,m), ans = 0;
	REP(i,1,mx) {
		int r1 = 0, r2 = 0;
		for (int j=i; j<=n; j+=i) r1+=mu[j];
		for (int j=i; j<=m; j+=i) r2+=mu[j];
		ans += mu2[i]*r1*r2;
	}
	printf("%d
", ans);
}

int main() {
	init();
	int t;
	scanf("%d", &t);
	while (t--) work();
}