SQFREE - Square-free integers

求n以内，约数中不包含任意一个平方数的个数，\(n≤10^{14}\)。

解

显然为约数计数问题，于是想办法转换为代数问题，不难列出

\[ans=\sum_{i=1}^n\mu^2(i)\]

没有gcd，不好反演，考虑容斥，则有

\[ans=\sum_{i=1}^{\sqrt{n}}\mu(i)[n/i^2]\]

以此即可解决问题，时间复杂度\(O(\sqrt{n})\)。

参考代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#define il inline
#define ri register
#define ll long long
#define Size 10000000
using namespace std;
bool check[Size+1];
int prime[1000000],pt,mu[Size+1];
template<class free>
il void read(free&);
il void prepare(int);
int main(){int lsy;read(lsy);ll n,i,j,ans;prepare(Size);while(lsy--){read(n),ans&=0;for(i=1;i*i<=n;i=j+1)j=sqrt(n/(n/(i*i))),ans+=(mu[j]-mu[i-1])*(n/(i*i));printf("%lld\n",ans);}return 0;
}
il void prepare(int n){ri int i,j;mu[1]=1;for(i=2;i<=n;++i){if(!check[i])prime[++pt]=i,mu[i]=-1;for(j=1;j<=pt&&prime[j]*i<=n;++j){check[i*prime[j]]|=true;if(!(i%prime[j]))break;mu[i*prime[j]]=-mu[i];}}for(i=1;i<=n;++i)mu[i]+=mu[i-1];
}
template<class free>
il void read(free &x){x&=0;ri char c;while(c=getchar(),c<'0'||c>'9');while(c>='0'&&c<='9')x=(x<<1)+(x<<3)+(c^48),c=getchar();
}

转载于:https://www.cnblogs.com/a1b3c7d9/p/10884606.html

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