中国民生银行信用卡中心金融科技挑战赛数据算法赛道面经
TimeLine：线上初赛202208??，线下复赛20220822-20220826
当时的BG：北邮本硕，管理类专业，两段实习经历：字节数据分析师、美团商业分析师
写在前面的话：以下问题的答案必然存在错误，敬请读者批评指正
线上初赛
主要考察算法和SQL，算法的难度不是特别高，但SQL有一定难度
民生银行卡中心的代码平台仅有上传功能，并不是OJ，因此只能在本地进行样例测试
初试没有屏幕监控，因此理论上可求助于任何渠道（划掉）
算法题
1.
题目描述：
输入格式：
输出格式：
input_nums = []
while True:
inputing = int(input())
if inputing == -1:
break
else:
input_nums.append(inputing)
res = [0] * len(input_nums)
for index in range(len(input_nums)):
for i in range(1, input_nums[index] + 1):
res[index] += input_nums[index]//i * i
for num in res:
print(num)
input_nums = []
while True:
inputing = int(input())
if inputing == -1:
break
else:
input_nums.append(inputing)
res = [0] * len(input_nums)
for index in range(len(input_nums)):
for i in range(1, input_nums[index] + 1):
tmp = []
for j in range(1, int(i ** 0.5) + 1):
if i % j == 0:
tmp.append(j)
tmp.append(int(i/j))
res[index] += sum(set(tmp))
for num in res:
print(num)
2.
题目描述：
输入格式：
输出格式：
def isprime(x):
if x < 2:
return False
for i in range(2, int(x ** 0.5) + 1):
if x % i == 0:
return False
return True
input_nums = []
while True:
inputing = int(input())
if inputing == -1:
break
else:
input_nums.append(inputing)
res = [0] * len(input_nums)
for index in range(len(input_nums)):
for i in range(1, input_nums[index] + 1):
tmp = []
if (int(str(i)) != int(str(i)[::-1])) and isprime(i) and isprime(int(str(i)[::-1])):
tmp.append(i)
res[index] += sum(tmp)
for num in res:
print(num)
3.
题目描述：
输入格式：
输出格式：
input_nums = []
while True:
inputing = int(input())
if inputing == -1:
break
else:
input_nums.append(inputing)
res = [0] * len(input_nums)
for index in range(len(input_nums)):
for i in range(1, input_nums[index] + 1):
res[index] += str(i).count('2')
for num in res:
print(num)
SQL题
注：SQL题在本地不好测试，因此以下代码仅提供思路上的参考
1.
数据表详情如下：
数据表示例性字段如下：
问题1：请用sql查询语句，给出login_day中不同日期的种类个数以及最大日期和最小日期是什么？
select count(login_day) as '日期种类', max(login_day) as '最大日期', min(login_day) as '最小日期' from app_data group by login_day
问题2：连登次数是指连续登录success且没有出现timeout或fail，login_day不需要连续。请写一段sql语句来计算每个客户最多的连登次数，结果以client_id升序排列
with a as
(
select client_id,
login_day,
result,
lead(client_id, 1) over(partition by client_id order by login_day) as next_result
from app_data
)
select client_id,
count(result) as '次数'
from a
where result = 'success' and next_result = 'success'
group by client_id
order by client_id asc
问题3：请用sql语句来计算每个客户最多的连续非fai的次数（login_day不需连续），结果以client_id升序排列
with a as
(
select client_id,
login_day,
result,
lead(client_id, 1) over(partition by client_id order by login_day) as next_result
from app_data
)
select client_id,
count(result) as '最多连续非fail次数'
from a
where result != 'fail' and next_result != 'fail'
group by client_id
order by client_id asc
问题4：连续登录是指login_day连续不间断。请写出一段sql语句来计算每个客户连续登录success大于等于4天的次数，结果以client_id升序排列
with a as
(
select client_id,
login_day,
result,
lead(client_id, 1) over(partition by client_id order by login_day) as next_result
from app_data
)
select client_id,
count(result) as '次数'
from a
where result = 'success' and next_result = 'success'
group by client_id
having count(result) >= 4
order by client_id asc
2.
数据表详情如下：
数据表示例性字段如下：
问题1：请用sql语句计算首次交易后第二天再次交易的用户的比率，四舍五入到小数点后两位
with a as
(
select client_id, trans_day,
row_number() over(partition by client_id order by trans_day asc) as rn,
date_sub(trans_day, interval rn day) as sub
from trans_data
where rn in (1, 2)
)
select round(count(client_id) / (select count(client_id) from trans_data group by client_id), 2) as Ratio
from a
group by client_id
having count(distinct sub) = 1
问题2：请用sql语句查询每个客户最近一次交易的amount，将amount按降序排列并给出前20行
with a as
(
select client_id,
trans_day, amount,
row_number() over(partition by client_id order by trans_day desc) as rn
from trans_data
)
select client_id, trans_day as '最近交易日期', amount
from a
where rn = 1
order by amount desc
limit 20
线下复赛
共有40名同学进入线下复赛环节，其中研发赛道20名，数据算法赛道20名
每5人为1个小组，因此数据算法赛道共有4支队伍，每两支队伍共享一个复赛题目，题目内容与银行风控有关
每支队伍需要在给定时间内，根据复赛的试题材料，进行线上打榜比赛，并制作PPT回答材料相关问题
复赛时间为5天，较招行夏令营而言相对不那么累，奖励较为丰厚，有一半的队伍可获得数目不菲的奖金，因此相当推荐大家参与
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