[bzoj] 1101 Zap || 莫比乌斯反演

原题

求x<=a，y<=b，并且gcd(x,y)=d的x，y对数

---

(sum^{n}_{x=1}sum^{m}_{y=1}[gcd(x,y)==d])
(=sum^{lfloor frac{n}{d} floor}_{x=1}sum^{lfloor frac{m}{d} floor}_{y=1}[gcd(x,y)==1])
根据莫比乌斯反演，可以把式子化为
(=sum^{lfloor frac{n}{d} floor}_{x=1}sum^{lfloor frac{m}{d} floor}_{y=1}sum_{d|gcd(x,y)}mu(d))
因为$d|gcd(x,y) $ => (d|x,d|y)
所以式子变为 (sum^{n/d}_{i=1}mu(d)*lfloor frac{lfloor frac{n}{d} floor}{d} floor*lfloor frac{lfloor frac{m}{d} floor}{d} floor)

用n/d的取值分一个段，然后对mu求一个前缀和即可解决……

分段：
因为我们知道(lfloor frac{n}{d} floor)的取值有(sqrt n)种取值，所以对于(lfloor frac{n}{d} floor)和(lfloor frac{m}{d} floor)一共有(sqrt n + sqrt m)种情况
经典枚举方法：

for (int i=1,last=0;i<=a;i=last+1)
	{
	    last=min(a/(a/i),b/(b/i));
	    ans+=(ll)(a/i)*(b/i)*(sum[last]-sum[i-1]);
	}

对mu求前缀和：
利用欧拉筛法：

void init()
{
    miu[1]=sum[1]=1;
    for (int i=2;i<=N;i++)
    {
	if (!notprime[i]) miu[i]=-1,prime[++tot]=i;
	for (int j=1;j<=tot && i*prime[j]<=N;j++)
	{
	    notprime[i*prime[j]]=1;
	    if (i%prime[j]) miu[i*prime[j]]=-miu[i];
	    else
	    {
		miu[i*prime[j]]=0;
		break;
	    }
	}
	sum[i]=sum[i-1]+miu[i];
    }
}

AC代码：

#include<cstdio>
#include<algorithm>
#define N 50000
typedef long long ll;
using namespace std;
int t,a,b,d,miu[N+10],prime[N+10],tot,sum[N];
bool notprime[N+10];
ll ans;

int read()
{
    int ans=0,fu=1;
    char j=getchar();
    for (;j<'0' || j>'9';j=getchar()) if (j=='-') fu=-1;
    for (;j>='0' && j<='9';j=getchar()) ans*=10,ans+=j-'0';
    return ans*fu;
}

void init()
{
    miu[1]=sum[1]=1;
    for (int i=2;i<=N;i++)
    {
	if (!notprime[i]) miu[i]=-1,prime[++tot]=i;
	for (int j=1;j<=tot && i*prime[j]<=N;j++)
	{
	    notprime[i*prime[j]]=1;
	    if (i%prime[j]) miu[i*prime[j]]=-miu[i];
	    else
	    {
		miu[i*prime[j]]=0;
		break;
	    }
	}
	sum[i]=sum[i-1]+miu[i];
    }
}

int main()
{
    init();
    t=read();
    while (t--)
    {
	a=read();b=read();d=read();
	a/=d;
	b/=d;
	ans=0;
	if (a>b) swap(a,b);
	for (int i=1,last=0;i<=a;i=last+1)
	{
	    last=min(a/(a/i),b/(b/i));
	    ans+=(ll)(a/i)*(b/i)*(sum[last]-sum[i-1]);
	}
	printf("%lld
",ans);
    }
    return 0;
}