CF-60D - Savior(并查集+数论)

D - Savior

Time Limit:4000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

Misha decided to help Pasha and Akim be friends again. He had a cunning plan — to destroy all the laughy mushrooms. He knows that the laughy mushrooms can easily burst when they laugh. Mushrooms grow on the lawns. There are a[t] mushrooms on the t-th lawn.

Misha knows that the lawns where the mushrooms grow have a unique ability. A lawn (say, i) can transfer laugh to other lawn (say, j) if there exists an integer (say, b) such, that some permutation of numbers a[i], a[j] and b is a beautiful triple (i ≠ j). A beautiful triple is such three pairwise coprime numbers x, y, z, which satisfy the following condition: x² + y² = z².

Misha wants to know on which minimal number of lawns he should laugh for all the laughy mushrooms to burst.

Input

The first line contains one integer n (1 ≤ n ≤ 10⁶) which is the number of lawns. The next line contains n integers a_i which are the number of mushrooms on the i-lawn (1 ≤ a_i ≤ 10⁷). All the numbers are different.

Output

Print a single number — the minimal number of lawns on which Misha should laugh for all the mushrooms to burst.

Sample Input

Input

1
2

Output

1

Input

2
1 2

Output

2

Input

2
3 5

Output

1

思路：所有勾股数都可以表示为x^2-y^2,,2*x*y,,,x^2+y^2  枚举所有勾股数，然后把能合并的合并，最后查看有几个不同的集合就是答案。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int mm=1e7+9;
int root[mm],f[mm],n,num;
int look(int x)
{
  if(x^root[x])
    root[x]=look(root[x]);
  return root[x];
}
void uni(int a,int b)
{
  a=look(a);b=look(b);
  if(root[a]!=0&&root[b]!=0&&a!=b)
  {
    root[a]=b;++num;
  }
}
int gcd(int a,int b)
{ int z;
  while(a)
  {
    z=a;a=b%a;b=z;
  }
  return b;
}
int main()
{ long long x,y,z;
  while(scanf("%d",&n)!=EOF)
    { memset(root,0,sizeof(root));
      for(int i=0;i<n;++i)
        {scanf("%d",&f[i]);root[f[i]]=f[i];}
        num=0;
      for(long long i=1;i<mm;i++)
      {
        for(long long j=i+1;2*i*j<mm;j+=2)
        {
          x=j*j-i*i;y=2*i*j;z=i*i+j*j;
          if(x>mm||y>mm)break;
          if(gcd(i,j)!=1)continue;
            uni(x,y);
          if(z<mm)
            uni(x,z);
          if(z<mm)
            uni(y,z);
        }
      }
      printf("%d\n",n-num);
    }
}