[data mining engineer - written examination] Haier company in 2022

company ： haier

Data mining engineer

1 choice question （18 individual ）

Can learn question bank ：https://brainly.in/
Esmander's logic problem , Note that only 20 minute , There is no time prompt , I didn't finish it. It's time to finish it

（1）5:124::7:？

answer ：342
analysis ：$5^3$-1 = 124 , and $7^3$ -1=342

（2） G3S:J3P::L4X:?

answer ：P4T
analysis ： In alphabetical order ,L + 4 = P （L It's No 12 Letter , P The first 16 Letters ）.X - 4 = T （X It's No 24 A title , T The first 20 Letters ）

（3） Which of the following options is different from others

A. NUTA
B. REOS
C. RREBHTO
D. RISSTE
answer ：REOS
analysis ：
Reorder letters , Classify according to the meaning of words
NUTA = AUNT（ aunt ）
REOS = ROSE ( Rose )
RREBHTO = BROTHER（ brother ）
RISSTE = SISTER（ sister ）
Only aunts are not relatives

（4）82,97,114,133,？

answer ：154
analysis ： Is a sequence of differences between two numbers , Is an isochronous sequence , The difference is 2
97-82 = 15
114-97= 17
133 - 114 = 19
Difference between 2, that 133+19+2 = 154

（5） Encryption method ：MATH The corresponding alphabet is RFYM, that PHYSICS The corresponding letter

answer ：UMDXNHX
analysis ：
M—R (M Turn into R, stay M After that 5 Letters )
A——F (A become F, It's also A The last one 5 Letters )
T—Y (T become Y, Y It's also T The last one 5 Letters )
H—M (H Turn into M, It's also H Later 5 Letters )
therefore ,PHYSICS Will be encoded as UMDXNHX
P—U (U In the 5 Letters P after )
H—M (M It's also H Later 5 Letters )
Y - D (Y It's also Y The last one 5 Letters , Because here we have to count again )
S—X (X It's also S Later 5 Letters )
I - N (N Also in I After that 5 Letters )
C—H (H It's also C After that 5 Letters )
S—X (X It's also S Later 5 Letters )
therefore ,PHYSICS Will be encoded as UMDXNHX.

（6）AB Two people have the same starting point ,A Run east 4 km ,B Run west 3 km ,A Run to the left 4 km ,B Run to the right hand 4 km , Ask at this time , How far apart .

answer ：7

Other omitted ..

2 Elementary programming problem （2 individual ）

45 minute
（1） Calculate the sum of odd numbers in a number

def countNum(self,n):
	num = list(str(n))
	result = 0
	for i in num:
		if int(i)%2!=0:
			result +=int(i)
	return result
		

（2） Simulated signal transmission , Send string A, Receive string B, Judge whether the two signals are consistent , In case of disagreement , Output the wrong character
Example ：

Input ：abcdefgh abcdefg
Output ：h

Input abcd abcd
Output ：NA

def judge(self,s)
	st = s.split(' ')
	s1 = st[0]
	s2 = st[1]
	p,q = len(s1)-1,len(s2)-1
	tag = 0
	if len(s1)==len(s2):
		print('NA')
	while p>=0 and q>=0:
		if s1[p] !=s2[q]:
			tag = 1
			print(s1[p])
			break
	if tag==0:
		print(s1[-1])