bzoj2986 Non-Squarefree Numbers （莫比乌斯函数）

来源：互联网发布：java发牌程序编辑：程序博客网时间：2024/06/11 04:14

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2986
首先能想到二分答案，然后判断1~X中有多少个squarefree，这里用莫比乌斯函数来容斥即可。

#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<iostream>#include<algorithm>#include<set>#include<map>#include<queue>#include<stack>#include<vector>#include<bitset>using namespace std;typedef long long LL;int miu[510000];int prim[510000],primm;bool valid[510000];void mobius(int N){    miu[1]=1;    for (int i=2;i<=N;i++)    {        if (!valid[i]) prim[++primm]=i,miu[i]=-1;        for (int j=1;j<=primm&&i*prim[j]<=N;j++)        {            valid[i*prim[j]]=1;            if (i%prim[j]==0)            {                miu[i*prim[j]]=0;                break;            }            else miu[i*prim[j]]=-miu[i];        }    }}LL cal(LL x){    LL res=0;    for (LL i=1;i*i<=x;i++)        res+=x/i/i*(LL)miu[i];    return x-res;}LL ans,n;int main(){    cin>>n;    mobius(500000);    LL l=1,r=210000000000LL;    while (l<=r)    {        LL mid=(l+r)/2;        if (cal(mid)>=n) r=mid-1,ans=mid;        else l=mid+1;    }    cout<<ans<<endl;    return 0;}

0 0