bzoj2820:YY的GCD

传送门

莫比乌斯反演，算是道模板题吧，但是比[POI2007]Zap难一些，zap我也有题解

对于这个题，一贯的套路，我们设

\[f(d)=\sum_{i=1}^{a}\sum_{j=1}^{b}[gcd(i,j)==d]\\ g(n)=\sum_{n|d}f(d)=\sum_{i=1}^{a}\sum_{j=1}^{b}[n|gcd(i,j)]=\lfloor \frac{a}{n}\rfloor\lfloor \frac{b}{n} \rfloor \]

然后反演

\[f(d)=\sum_{d|n}\mu(\frac{n}{d})g(n) \]

然后考虑答案

\[ans=\sum_{d\in prime}\sum_{d|n}\mu(\frac{n}{d})g(n)\\ ans=\sum_{d\in prime}\sum_{d|n}\mu(\frac{n}{d})\lfloor \frac{a}{n}\rfloor\lfloor \frac{b}{n} \rfloor \]

设\(T=\frac{n}{d}\)

\[ans=\sum_{d\in prime}\sum_{T=1}^{min(a,b)/d}\mu(T)\lfloor \frac{a}{Td}\rfloor\lfloor \frac{b}{Td} \rfloor \]

再设\(k=Td\)

\[ans=\sum_{d\in prime}\sum_{k=1}^{min(a,b)}\mu(\frac{k}{d})\lfloor \frac{a}{k}\rfloor\lfloor \frac{b}{k} \rfloor\\ ans=\sum_{k=1}^{min(a,b)}\sum_{d\in prime}\mu(\lfloor \frac{k}{d} \rfloor)\lfloor \frac{a}{k}\rfloor\lfloor \frac{b}{k} \rfloor\\ ans=\sum_{k=1}^{min(a,b)}\sum_{d\in prime,d|k}\mu(\frac{k}{d})\lfloor \frac{a}{k}\rfloor\lfloor \frac{b}{k} \rfloor\\ ans=\sum_{k=1}^{min(a,b)}\lfloor \frac{a}{k}\rfloor\lfloor \frac{b}{k} \rfloor(\sum_{d\in prime,d|k}\mu(\frac{k}{d}))\\ \]

然后后面可以线筛之后前缀和一下，剩下的数论分块就好了

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
void read(int &x) {
	char ch; bool ok;
	for(ok=0,ch=getchar(); !isdigit(ch); ch=getchar()) if(ch=='-') ok=1;
	for(x=0; isdigit(ch); x=x*10+ch-'0',ch=getchar()); if(ok) x=-x;
}
#define rg register
const int maxn=1e7;
int T,n,m,pri[maxn/10],tot,mu[maxn+10],g[maxn+10];
bool vis[maxn+10];long long ans,sum[maxn+10];
void prepare()
{
	mu[1]=1;
	for(rg int i=2;i<=maxn;i++)
	{
		if(!vis[i])pri[++tot]=i,mu[i]=-1;
		for(rg int j=1;j<=tot&&pri[j]*i<=maxn;j++)
		{
			vis[i*pri[j]]=1;
			if(!(i%pri[j]))break;
			else mu[i*pri[j]]=-mu[i];
		}
	}
	for(rg int i=1;i<=tot;i++)
		for(rg int j=1;j*pri[i]<=maxn;j++)g[j*pri[i]]+=mu[j];
	for(rg int i=1;i<=maxn;i++)sum[i]=sum[i-1]+g[i];
}
int main()
{
	read(T),prepare();
	while(T--)
	{
		read(n),read(m);if(n>m)swap(n,m);ans=0;
		for(rg int i=1,j;i<=n;i=j+1)
		{
			j=min(n/(n/i),m/(m/i));
			long long t=1ll*(n/i)*(m/i)*(sum[j]-sum[i-1]);
			ans+=t;
		}
		printf("%lld\n",ans);
	}
}