题解:

莫比乌斯反演

ans=sigma(x/(i*i)*miu[i])

代码:

#include<bits/stdc++.h>
using namespace std;
const int N=44723;
int T,n,flag[N],p[N],tot,miu[N];
void init()
{
    miu[1]=1;
    for (int i=2;i<N;i++)
     {
         if (!flag[i])
          {
              miu[i]=-1;
              p[++tot]=i;
          }
         for (int j=1;j<=tot;j++)
          {
              int k=p[j]*i;
              if (k>=N)break;
              flag[k]=1;
              if (i%p[j]==0)
               {
                   miu[k]=0;
                   break;
               }
              miu[k]-=miu[i];
          }
     }
}
int pd(int x)
{
    int ans=0;
    for (int i=1;i<=sqrt(x);i++)ans+=miu[i]*(x/(i*i));
    return ans>=n;
}
int main()
{
    scanf("%d",&T);
    init();
    while (T--)
     {
         scanf("%d",&n);
        int l=0,r=2*n;
        while (l<r)
         {
             int mid=((long long)l+r+1)/2;
             if (!pd(mid))l=mid;
             else r=mid-1;
         }
        printf("%d\n",l+1); 
     }
}

 

posted on 2018-03-10 13:59  宣毅鸣  阅读(94)  评论(0编辑  收藏  举报