当前时间:NOW() 前n天:DATE_SUB(NOW(),INTERVAL n DAY) 后n天:DATE_SUB(NOW(),INTERVAL -n DAY) 取前n条记录:SELECT * FROM 表名 LIMIT n 从第n条开始取m条:SELECT * FROM 表名 LIMIT n,m 注释:n从0开始计
前几年 select to_char(sysdate, 'yyyy') - level + 1 years from dual connect by level <= num num:即想获取几年的 后几年 select to_char(sysdate, 'yyyy') + level - 1 years from dual connect by level <=num num:即想获取几年的
1282 - Leading and Trailing You are given two integers: n and k, your task is to find the most significant three digits, and least significant three digits of nk. Input Input starts with an integer T (≤ 1000), denoting the number of test cases. Each
public class Test { public static void main(String[] args) { int age = 6; //先自加,再使用(age先自加1,然后再打印age=7,此时age的值在内存中是7) System.out.println("age=" + ++age); //先使用,再自加(此时age的值在内存中是7,先打印age=7,然后age再自加1,所以此时打印age=7,但是其实此时age的值在内存中已经是8了) System.out.pri
select * from aa01_2014 where aaa001=(select c.p from (select aaa001,lag(aaa001,1,0) over (order by aaa001) as p from aa01_2014) c where c.aaa001='8a9299ec522f54f401522f81eedc0007') ; select * from aa01_2014 where aaa001=(select c.n from (select aa
二叉树的遍历 Time Limit: 1000 MS Memory Limit: 32768 K Total Submit: 60(34 users) Total Accepted: 34(30 users) Rating: Special Judge: No Description 给出一棵二叉树的中序和前序遍历,输出它的后序遍历. Input 本题有多组数据,输入处理到文件结束. 每组数据的第一行包括一个整数n,表示这棵二叉树一共有n个节点. 接下来的一行每行包括n个整数,表示这棵树的中序遍
