List the prime numbers within 1000, and then use two numbers to represent the current number and the number he should multiply, ans+= the number of the number he should multiply, and the current number plus one
#include <bits/stdc++.h>
using namespace std;
int flag[1000006];
int prime[1009],N=0;
void init()
{
prime[N++]=2;
int j=2;
for(int i=3;i<=1007;i++)
{for(j=2;j*j<=i;j++)
{
if(i%j==0)break;
}
if(j*j>i)prime[N++]=i;
}
}
intmain()
{
int t;
init();
scanf("%d",&t);
while(t--)
{
int n,x;
scanf("%d",&n);
memset(flag,0,sizeof(flag));
int ans=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&x);
long long a=1,b=1;
int flag1=0;
for(int i=0;i<N&&prime[i]<=x;i++)
{
int tmp=prime[i]*prime[i]*prime[i];
while(x%tmp==0)x/=tmp;
if(x%(prime[i]*prime[i])==0)
{
a*=prime[i]*prime[i];
b*=prime[i];
x/=(prime[i]*prime[i]);
}
if(x%prime[i]==0)
{
a*=prime[i];
b*=prime[i]*prime[i];
x/=prime[i];
}
if(b>1e6)flag1=1;
if(flag1)break;
}
if(flag1)continue;
if(x>prime[N-1])continue;
ans+=flag[b];
//cout<<b<<" "<<flag[b]<<endl;
flag[a]++;
}
printf("%d\n",ans);
}
return 0;
}