# Python題目反思 / 建國中學_邱浚祐
### Ch4 串列：f071: 2. 刮刮樂 (Lottery)<br>
題目整理：<br>
1. input 3 lucky numbers(integer)<br>
2. input 5 numbers(integer)<br>
3. input 5 amount of money(integer)<br>
4. each number in (2.) is linked to the corresponding money amount in (3.)<br>
5. if previous two lucky number is equal to any number in (2.), we get the corresponding amount of money in (3.)<br>
6. however, if the third lucky number is equal to any number in (2.), we lose the corresponding amount of money in (3.), whereas we won't have negative number in total<br>
7. if the third lucky number is NOT equal to any number in (2.), we can get double amount of money in total<br>
8. output the money that we can get at the end<br>
```python
lucky=list(map(int, input().split())) #get lucky num
num=list(map(int, input().split())) #get num related to money
money=list(map(int, input().split())) #get lottery money
luck=True #bool for whether is able to be double up
sum=0 #total amount of monry we'll get
for i in range(2):
for k in range(5):
if lucky[i]==num[k]: #whether lucky num is in num zone
sum+=money[k] #if True, get corresponding money
for k in range(5):
if lucky[2]==num[k]: #whether THIRD luck num is in num zone
sum-=money[k] #if True, lose corresponding money
luck=False # , and will NOT be able to double up
if sum<0:
print("0") #NO NEGATIVE outputs
elif luck==True:
print(sum*2) #THIRD lucky num is NOT in num zone, double up
else:
print(sum) #THIRD lucky num is in num zone
```
6 25 39
34 6 17 25 29
500 400 300 600 900
2000
Further improvement:<br>
  use "in" and "index" to solve the problem<br>
  c.f. https://stackoverflow.com/questions/9542738/python-find-in-list
***
### Ch5. while迴圈：a149: 乘乘樂<br>
題目整理：<br>
1. get a number which tells how many times of input<br>
2. get an input number<br>
3. print the result that each digit in the number(2.) times each other<br>
```python
try:
while True:
T=int(input()) #get input times as T
for i in range(T): #repeat T times
n=input() #input num
sum=int(n[0]) #sum default as the first digit
for i in range(1,len(n)): #for start at second, then to the last digit
sum*=int(n[i]) #make sum times number i digit
print(sum) #print sum out
except EOFError: #EOF(not working in jupyter here)
pass
```
3
356
90
123
6
9999
6561
Further improvement:<br>
  use list(str) to solve the problem faster<br>
  c.f. https://stackoverflow.com/questions/4978787/how-to-split-a-string-into-a-list-of-characters
***
### 04/25 情書解密<br>
題目整理：<br>
1. reverse the fisrt half string and the second half
2. reverse the whole article
3. delete capitalized letters
4. turn "*" to "i", "2" to "to"
5. capitalized letter "I" if it represents first person subject, and turn "_" to " "(space)
6. capitalized the first letter that starts a sentence
```python
import re
s=input() #get the encrypted words
n=len(s)//2 #n=length of a half of article
chunks = [s[i:i+n] for i in range(0, len(s), n)] #set chunk as a list for two half, ref[1]
ans1=chunks[1]+chunks[0] #ans1=reverse the half split
ans2=''.join(reversed(ans1)) #reverse the whole article, ref[2]
ans3=re.sub(r'[A-Z]', '', ans2) #delete capitalized letters, ref[3]
ans4=ans3.replace("*", "i") #replace "*" with "i"
ans4=ans4.replace("2", "to") #replace "2" with "to"
ans5=ans4.replace(" i ", " I ") #replace "i" with "I" if a space is before and after the letter "i"
ans5=ans5.replace("_", " ") #replace "_" with " "
ans6=( '. '.join(map(lambda s: s.strip().capitalize(), ans5.split('.')))) #let first letter that starts a sentence be capitalized, ref[4]
ans6=ans6.replace(" i ", " I ") #the former line will let the "i" transfer not working so redo the "i" to "I"
print(ans6) #print the result
```
WnHCaKeUDmOWT_wYeQn_elohw_a_seTv*g_dnaIKGJ_em_Mserussaer_ed*s_ym_yb_gUn*yatMs_TuoyL_fLo_thgUuHoht_ehtYV.ed*s_ym_yb_uoy_hFt*w_ma_*Y_VetelOpmoc_dnVa_ypSKpah_wIoh_dZeRQz*laUer_QreNvePVKNn_*_.ef*l_PyFm_2J_OgKn*
The thought of you staying by my side reassures me and gives a whole new meaning to my life. I never realized how happy and complete I am with you by my side.
ref[1]: https://stackoverflow.com/questions/312443/how-do-i-split-a-list-into-equally-sized-chunks, <br>      https://medium.com/quick-code/python-one-liner-list-comprehension-24e35a20bd1d<br>
ref[2]: https://stackoverflow.com/questions/931092/reverse-a-string-in-python<br>
ref[3]: https://www.geeksforgeeks.org/python-eliminate-capital-letter-starting-words-from-string<br>
ref[4]: https://stackoverflow.com/questions/22800401/how-to-capitalize-the-first-letter-of-every-sentence
Sample Input #1:<br>
WnHCaKeUDmOWT_wYeQn_elohw_a_seTv\*g_dnaIKGJ_em_Mserussaer_ed\*s_ym_yb_gUn\*yatMs_TuoyL_fLo_thgUuHoht_ehtYV.ed\*s_ym_yb_uoy_hFt\*w_ma_\*Y_VetelOpmoc_dnVa_ypSKpah_wIoh_dZeRQz\*laUer_QreNvePVKNn_\*_.ef\*l_PyFm_2J_OgKn\*
Further improvement:<br>
  think a easier way to capitalize the letter that starts a sentence<br>
***
### 考前複習：i096<br>
題目整理：<br>
1. get the number of the stubent(while I wouldn't need this variable)
2. get the amount of lunch budget for each student
3. output how many students are more or equal to the average of budgets
```python
try:
while True:
n=int(input())
total=0
budget=list(map(int, input().split()))
for i in range(len(budget)):
if budget[i] >= (sum(budget)/len(budget)):
total+=1
print(total)
except EOFError:
pass
```
5
60 72 35 88 51
2
Further improvement:<br>
  use sum with for inside print to make code shorter, shown below:<br>
  c.f. https://stackoverflow.com/questions/10543303/number-of-values-in-a-list-greater-than-a-certain-number
```python
try:
while True:
n=int(input())
budget=list(map(int, input().split()))
print(sum(budget[i] >= (sum(budget)/len(budget)) for i in range(len(budget))))
except EOFError:
pass
```
5
60 72 35 88 51
2
***
2022.6.23
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