BZOJ 1101 [POI2007]Zap 莫比乌斯反演

题意:链接

方法:莫比乌斯反演?

解析:

题中求的就是∑1<=x<=a∑1<=y<=b(gcd(x,y)==d)\sum\limits_{1<=x<=a}\sum\limits_{1<=y<=b}(gcd(x,y)==d)

即可转化为∑1<=x<=a/d∑1<=y<=b/d(gcd(x,y)==1)\sum\limits_{1<=x<=a/d}\sum\limits_{1<=y<=b/d}(gcd(x,y)==1)

又因为∑i=1nμ(i)=n==1?1:0\sum\limits_{i=1}^{n}\mu(i)=n==1?1:0

所以原公式可以转化为∑1<=x<=a/d∑1<=y<=b/d∑d|(x,y)μ(d)\sum\limits_{1<=x<=a/d}\sum\limits_{1<=y<=b/d}\sum\limits_{d|(x,y)}\mu(d)

因为d|(x,y)d|(x,y)所以d|xd|x d|yd|y，那么我们可以把和式提前

即公式变为∑1<=d<=min(a/d,b/d)μ(d)∑1<=x<=a/d且d|x∑1<=y<=b/d且d|y\sum\limits_{1<=d<=min(a/d,b/d)}\mu(d)\sum\limits_{1<=x<=a/d且d|x}\sum\limits_{1<=y<=b/d且d|y}

根据约数研究那道题，原公式可以进一步转化为

∑1<=d<=min(a/d,b/d)μ(d)[a/dd][b/dd]\sum\limits_{1<=d<=min(a/d,b/d)}\mu(d)[\frac{a/d}{d}][\frac{b/d}{d}]

接下来就可以暴力分块了！

μ(d)\mu(d)部分维护个前缀和,剩下的因为[a/dd][\frac{a/d}{d}]与[b/dd][\frac{b/d}{d}]具有单调不上升性,所以只需要计算每一块就好了。

代码:

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 50010
using namespace std;
int tot,t;
int a,b,d;
int prime
;
bool f
;
int miu
;
int sum
;
void sieve()
{
    miu[1]=1;
    for(int i=2;i<=50000;i++)
    {
        if(!f[i])
        {
            prime[++t]=i;
            miu[i]=-1;
        }
        for(int j=1;j<=t&&i*prime[j]<=50000;j++)
        {
            f[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                miu[i*prime[j]]=0;
                break;
            }else miu[i*prime[j]]=-miu[i];
        }
    }
    for(int i=1;i<=50000;i++)
    {
        sum[i]=sum[i-1]+miu[i];
    }
}
int main()
{
    sieve();
    scanf("%d",&tot);
    for(int i=1;i<=tot;i++)
    {
        scanf("%d%d%d",&a,&b,&d);
        int a1=a/d,b1=b/d;
        int x=min(a1,b1);
        int pos,ans=0;
        for(int i=1;i<=x;i=pos+1)
        {
            pos=min((a1/(a1/i)),(b1/(b1/i)));
            ans+=(sum[pos]-sum[i-1])*(a1/i)*(b1/i);
        }
        printf("%d\n",ans);
    }
}