CF 955C Sad Power Enumeration (Perfect Powers)

Question meaning: ask Q times, each time ask how many x in [L, R] satisfy x=a^p (a>0,p>1)
Q<=1e5. 1<=L<=R<= 1e18.


The upper bound is 1e18, then p will not exceed 61 at most .
Now calculate how many x of f[1,n] satisfy x=a^p. Then the answer is f[r]-f[l-1
] Take p, what is the maximum a of a^p<=n at this time, and then find that repeated processing can't come....wa...

You can violently preprocess all Perfect Numbers within 1e18, p>=3. O(1e6).

Then for a given n, there are sqrt(n) of p==2. Then when p>=3, the last <=n position in the bisection in the vector can be used.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+5;
vector<ll> v;
void init()
{
	ll n=1e18;
	for(ll i=2;i*i*i<=n;i++) //when p>=3  enumerate a
	{
		ll s=i*i;
		while(s<=n/i)
		{
			s*=i;
			ll t=sqrt(s);
			if(t*t!=s) //
				v.push_back(s);
		}
	}
	sort(v.begin(),v.end());
	v.erase(unique(v.begin(),v.end()),v.end());
}
ll calc(ll n)
{
	ll idx=lower_bound(v.begin(),v.end(),n)-v.begin();	
	if(idx<v.size()&&v[idx]>n)
		idx--;
	if(idx==v.size())
		idx--;
	//cout<<n<<' '<<sqrt(n)<<' '<<idx<<'\n';
	return ll(sqrt(n))+idx;
}
intmain()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	init();
	ll Q,l,r;
	cin>>Q;
	while(Q--)
	{
		cin>>l>>r;
		cout<<calc(r)-calc(l-1)<<'\n';
	}
	return 0;
}