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~