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题解 | #截取出年龄#

SELECT SUBSTRING_INDEX( SUBSTRING_INDEX( `profile`, ',',- 2 ), ',', 1 ) AS age, COUNT(*) AS number FROM user_submit GROUP BY age

2023-08-05

0 248

题解 | #计算用户的平均次日留存率#

SELECT count( q2.device_id ) / count( q1.device_id ) AS avg_ret FROM ( SELECT DISTINCT device_id, date FROM question_practice_detail ) AS q1 LEFT ...

2023-08-04

0 303

题解 | #查看不同年龄段的用户明细#

select device_id,gender, case when age < 20 then '20岁以下' when age >= 20 and age <= 24 then '20-24岁' when age >=...

2023-08-03

0 251

题解 | #计算25岁以上和以下的用户数量#

select '25岁以下' as age_cut, count(distinct device_id) as 'number' from user_profile where age < 25 or age is null union select...

2023-08-03

0 228

题解 | #Old Bill#

#include<stdio.h> int main() { int n,x,j,y,z,p,q; while(scanf("%d",&n) != EOF) { scanf("%d %d %d",&x,&y,&z); ...

C++

2021-12-11

0 341

题解 | #与7无关的数#

#include<stdio.h> bool move(int i) { while(i) { if(i % 10 == 7) { return true; } i /= 10; } ret...

C++

2021-12-05

0 330

题解 | #对称平方数1#

#include<stdio.h> int main() { for(int i = 0;i <= 256 ; i++) { int j = i * i, sum = 0; while(j) { sum = sum ...

C++

2021-12-05

2 372

题解 | #反序数#

#include<stdio.h> int main() { int i, k ; for(i = 1000 ;i <= 9999;i++) { int j = i,sum = 0;//sum的值每次必须得更新，因为每次遍历的四位数是不同的 ...

C++

2021-12-05

0 348