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题解 | #kiki算数#

a,b=list(map(int,input().split())) if a+b>=100: if (a+b)%100//10==0: print((a+b)%10) else: print((a+b)%100) else: print...

Python3

2022-04-04

0 312

题解 | #小乐乐定闹钟#

time,spend=input().split() hour=int(time.split(':')[0]) minute=int(time.split(':')[1]) now_spend=hour*60+minute+int(spend) hour=now_spend//60%24 minut...

Python3

2022-04-04

0 308

题解 | #小乐乐定闹钟#

time,spend=input().split() hour=int(time.split(':')[0]) minute=int(time.split(':')[1]) now_spend=hour*60+minute+int(spend) hour=str(now_spend//60%24) ...

Python3

2022-04-04

0 288

题解 | #某乎问答最大连续回答问题天数大于等于3天的用户及其对应等级#

题目有个要求是最大连续回答问题天数大于等于3天，那按题意应该取一次max，留max_dyas_cnt作为days_cnt。事实上其实只选出days_cnt>=3也没关系，当然是属于最大连续回答问题天数大于等于3的用户，但是如果同一用户有两个这种连续记录那么就会重复（本题没有出现这种情况）,那...

Mysql

2022-03-05

0 372

题解 | #牛客直播各科目出勤率#

select course_id,course_name,round(cnt1/cnt2*100,2) as attend_rate from ( select course_id, course_name, count(distinct if(timestampdiff(se...

Mysql

2022-03-05

0 285

题解 | #某宝店铺连续2天及以上购物的用户及其对应的天数#

select user_id,count(distinct dt,dt1) as days_count from ( select user_id,dt,date_sub(dt,interval ranking day) as dt1 from ( select user_id, ...

Mysql

2022-03-04

0 276

题解 | #某宝店铺动销率与售罄率#尤其简单明了不易出错版！

select style_id, round(sum(item_sum)/sum(inventory-item_sum)*100,2) as pin_rate, round(sum(sales_price_sum)/sum(tag_price*inventory)*100,2) as...

Mysql

2022-03-04

0 264

题解 | #店铺901国庆期间的7日动销率和滞销率#

with t as ( select product_id,order_id,date(event_time) as dt,shop_id from tb_order_detail #order_id is not null，左连接选出的店铺和product_id都是有销量的 left jo...

Mysql

2022-03-04

0 327

题解 | #10月的新户客单价和获客成本#尤其通俗易懂版！结构化逻辑清楚

select round(avg(total_amount),1) as avg_amount, round(avg(price-total_amount),1) as avg_cost from ( select uid,order_id,total_amount,price,date...

Mysql

2022-03-04

0 286

题解 | #零食类商品中复购率top3高的商品#

select product_id, round(count(if(repurchase_cnt>1,uid,null))/count(uid),3) as repurchase_rate from ( select product_id,uid, count(if(dat...

Mysql

2022-03-04

0 292