传送门
思路:不断的分解质因子即可,最后注意 n 可能不为1,所以最后要记得乘n。
import java.math.BigInteger;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
long n = cin.nextLong();
long t = n;
int flag = 0;
long ans = 1;
for (long i = 2; i * i <= n; i++) {
flag = 1;
while (n % i == 0) {
n = n / i;
if (flag == 1) {
ans *= i;
flag = 0;
}
}
}
System.out.println(ans * n);
}
}