# Write your MySQL query statement below
select project_id,round(sum(experience_years)/count(project_id),2) as average_years
from
Project as p
left join
Employee as e
on p.employee_id=e.employee_id
group by project_id
/**
* 要执行的算法,返回结果v
*/
public interface Computable<A, V> {
public V comput(final A arg);
}
/**
* 用于缓存数据
*/
public class Memoizer<A, V> implements Computable<A,