思路1：二分+快速幂

#include<bits/stdc++.h>
using namespace std;
#define int long long
int n,m;
bool check(int a,int b){
	int ans=1;
	while(b){
		if(a>n)return false;
		if(b&1)ans*=a;
		if(ans>n)return false;
		a=a*a;
		b>>=1;
	}
	return ans<=n;
} 
void solve() {
	cin>>n>>m;
    int l=1,r=n+1;
    while(l+1<r){
    	int mid=l+r>>1;
    	if(check(mid,m))l=mid;
    	else r=mid;
	}
	cout<<l<<"\n";
}
signed main() {

	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	int tt=1;
	//cin>>tt;
	while(tt--) solve();
	return 0;
}
//3

思路2：

#include<bits/stdc++.h>
using namespace std;
#define int long long
int n,m;
void solve() {
	cin>>n>>m;
    int ans=pow(n,1.0/m);
    cout<<(int)ans;
}
signed main() {

	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	int tt=1;
	//cin>>tt;
	while(tt--) solve();
	return 0;
}
//3

 over~