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过去一年提交了

勋章 ①金银铜:在竞赛中获得第一二三名;②好习惯:自然月10天提交;③里程碑:解决1/2/5/10/20/50/100/200题;④每周打卡挑战:完成每周5题,每年1月1日清零。

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评论笔记

评论日期 题目名称 评论内容 站长评论
2025-01-08 大于J小于K的手牌 
输出示例直接把答案贴上来了。
尴尬了哈哈哈 已修改

提交记录

提交日期 题目名称 提交代码
2025-02-05 与X轴有且只有一个交点的一元二次函数 
select * from numbers_for_fun
where a!=0 and b*b-4*a*c = 0
order by id;
2025-02-05 开口向上且经过原点的一元二次函数 
select * from numbers_for_fun
where a >0 and c=0
order by id;
2025-02-05 开口向上的一元二次函数 
select * from numbers_for_fun 
where a >0 
order by id;
2025-02-05 必过(-1, 0)的一元一次函数 
select * from numbers_for_fun
where a = 0 and b != 0 and c = b
order by id;
2025-02-05 必过(3, -8)的一元一次函数 
select * from numbers_for_fun
where a = 0 and b != 0 and (3*b+c) = -8
order by id;
2025-02-05 必过(-1, -1)的一元一次函数 
select * from numbers_for_fun 
where a = 0 and b != 0 and c-b = -1
order by id;
2025-02-05 必过(0, 1)的一元一次函数 
select * from numbers_for_fun
where a = 0 and b !=0 and c =1
order by id;
2025-02-05 不经过第二象限的一元一次函数 
select * from numbers_for_fun 
where a = 0 and b >0 and c <= 0
order by id;
2025-02-05 不经过第三象限的一元一次函数 
select * from numbers_for_fun 
WHERE a = 0 and b <= 0 and -c/b >= 0
order by id;
2025-02-05 找出与y=x有交点的所有一元一次函数 
select * from numbers_for_fun
where a = 0 and b != 1 and b !=0
order by id;
2025-02-05 找出与y=x有交点的所有一元一次函数 
select * from numbers_for_fun
where a = 0 and b != 1
order by id;
2025-02-05 找出与X轴交点小于等于0的一元一次函数 
select * from numbers_for_fun 
where a = 0 and b*c>=0 and b != 0
order by id;
2025-02-05 找出与X轴交点小于等于0的一元一次函数 
select * from numbers_for_fun 
where a = 0 and b*c>0
order by id;
2025-02-05 找出与X轴交点大于0的一元一次函数 
select * from numbers_for_fun 
where a = 0 and b*c < 0 and b<>0
order by id asc;
2025-02-05 找出与X轴交点大于0的一元一次函数 
select * from numbers_for_fun 
where a = 0 and c >0 and b != 0
order by id asc;
2025-02-05 找出所有经过原点的一元一次函数 
select * from numbers_for_fun 
where a = 0 and c = 0 and b !=0
order by id asc;
2025-02-05 找出所有一元一次函数 
select * 
from numbers_for_fun
where a = 0 and b != 0
order by id asc;
2025-02-05 找出所有一元一次函数 
select * 
from numbers_for_fun
where a = 0
order by id asc;
2025-01-16 销售金额前10的商品信息 
select goods_id,sum(order_gmv) total_gmv
from order_info 
where date( order_time) = '2024-9-10'
group by goods_id
order by total_gmv desc
limit 10;
2025-01-16 抖音面试真题(1)T+1日留存率 
WITH recent_logins AS (
SELECT
usr_id,
DATE(login_time) AS login_date
FROM
user_login_log
WHERE
login_time >= DATE_SUB(CURDATE(), INTERVAL 30 DAY)
),
daily_users AS (
SELECT
login_date,
COUNT(DISTINCT usr_id) AS daily_user_count
FROM
recent_logins
GROUP BY
login_date
),
retained_users AS (
SELECT
r1.login_date AS login_date,
COUNT(DISTINCT r2.usr_id) AS retained_user_count
FROM
recent_logins r1
JOIN
recent_logins r2
ON
r1.usr_id = r2.usr_id
AND r2.login_date = DATE_ADD(r1.login_date, INTERVAL 1 DAY)
GROUP BY
r1.login_date
)
SELECT
du.login_date,
CONCAT(
ROUND(
IFNULL(ru.retained_user_count / du.daily_user_count * 100, 0),
2
),
'%'
) AS T1_retention_rate
FROM
daily_users du
LEFT JOIN
retained_users ru
ON
du.login_date = ru.login_date
ORDER BY
du.login_date;