Codeforces Round #596 div2 D. Power Products
\(预先枚举1e10以内所有数的k次方,然后每一个ai都去找他所有的1e5以内的倍数,更新答案。\)
\(复杂度\sqrt[k]{1e10}\times{\sqrt{1e5}\times{2}}\)
#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int N=1e5+100;
int n,k;
int a[N],vis[N];
ll b[N];
ll qpow(ll a,ll b)
{
ll ans=1;
while(b)
{
if(b&1)ans=ans*a;
if(ans>1e10||a>1e5)return 1e10+5;
a=a*a;
b>>=1;
}
return ans;
}
ll solve(int x)
{
int cnt=0;
ll res=1;
for(int i=2;i*i<=x;i++)
{
cnt=0;
while(x%i==0)
{
cnt++;
x/=i;
}
if(cnt)res*=qpow(i,k-((cnt-1)%k+1));
if(res>1e5+10)return res;
}
if(x!=1)res*=qpow(x,k-1);
return res;
}
int main()
{
//solve(40);
scanf("%d%d",&n,&k);
int cnt=0;
for(int i=1;;i++)
{
ll p=qpow(i,k);
if(p>1e10)break;
b[++cnt]=p;
}
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
ll ans=0;
for(int i=1;i<=n;i++)
{
ll p=solve(a[i]);
//cout<<p<<endl;
for(int j=1;j<=cnt;j++)
{
if(p*b[j]>1e5+10)break;
if(vis[p*b[j]])ans+=vis[p*b[j]];
//cout<<p*a[j]<<" "<<ans<<" "<<vis[p*b[j]]<<endl;
}
vis[a[i]]++;
}
printf("%lld\n",ans);
return 0;
}