#Z1643. 轻拍牛头

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一，86分解法

#include <bits/stdc++.h>
using namespace std;
long long n,ton[10000001],maxn;
struct nu
{
  long long id,par;
} niu[10000001],tn[1000001];
bool cmp(nu a,nu b)
{
  if(a.par == b.par) return a.id < b.id;
  return a.par < b.par;
}
int main()
{
  scanf("%lld",&n);
  for(int i = 1; i <= n; i++)
  {
    niu[i].id = i;
    scanf("%lld",&niu[i].par);
    tn[i] = niu[i];
    maxn = max(maxn,niu[i].par);
  }
  sort(niu + 1,niu + n + 1,cmp);
  for(int i = 1; i < n; i++)
  {
    int t = niu[i].par,j;
    if(niu[i].par == niu[i + 1].par) j = 1;
    else j = 2;
    while(1)
    {
      t *= j;
      if(t > maxn) break;
      ton[t]++;
      t /= j;
      j++;
    }
  }
  for(int i = 1; i <= n; i++) printf("%lld\n",ton[tn[i].par]);
  return 0;
}

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二，AC解法

#include <bits/stdc++.h>
using namespace std;
int n,tong[1000010],in[1000010],ans[1000010],ma;
int main()
{
  scanf("%d",&n);
  for(int i = 1; i <= n; i++)
  {
    scanf("%d",&in[i]);
	tong[in[i]]++;	
  	ma = max(ma,in[i]);
  }
  for(int i = 1; i <= ma; i++)
    if(tong[i] != 0)
      for(int j = i; j <= ma; j += i)
        ans[j] += tong[i];
  for(int i = 1; i <= n; i++) printf("%d\n",ans[in[i]] - 1);
  return 0; 
}