loop 10 question answer
第 1 題：判斷相加後是否為迴文
python
import sys
for line in sys.stdin.read().splitlines():
a, b = map(int, line.split())
s = str(a + b)
print(s, s == s[::-1])第 2 題：重複文字只保留一個
python
s = input()
# 最早出現的保留
seen = set()
first = ""
for ch in s:
if ch not in seen:
seen.add(ch)
first += ch
# 最晚出現的保留
seen = set()
last = ""
for ch in reversed(s):
if ch not in seen:
seen.add(ch)
last = ch + last
print(first)
print(last)第 3 題：密碼強度等級判斷
python
import re
def password_strength(pw):
score = 0
if len(pw) >= 8:
score += 1
if re.search(r'[A-Za-z]', pw) and re.search(r'\d', pw):
score += 1
if re.search(r'[A-Z]', pw):
score += 1
if re.search(r'[~!@#$%^&_()\[\]{}<>.,+\\/-]', pw):
score += 1
return ["Week", "Normal", "Nice", "Good", "Strong"][score]
try:
while True:
pw = input().strip()
if pw == "":
break
print(password_strength(pw))
except EOFError:
pass第 4 題：連續數字的最長子字串長度
python
import re
s = input()
numbers = re.findall(r'\d+', s)
if numbers:
print(max(len(num) for num in numbers))
else:
print(0)第 5 題：賣鴨子倒推
python
def duck(n, remain, steps=[]):
if n == 0:
print("出發時共", remain, "隻")
return remain
sold = remain + 1
total = sold * 2
steps.append((n, sold, remain))
result = duck(n - 1, total, steps)
if n == len(steps):
for i in reversed(steps):
print(f"經過第 {i[0]} 個村莊 賣 {i[1]} 隻,剩 {i[2]} 隻鴨")
return result
n, m = map(int, input().split())
duck(n, m)第 6 題：巴斯卡三角形
python
def pascal(n):
if n == 1:
return [[1]]
else:
triangle = pascal(n - 1)
prev_row = triangle[-1]
new_row = [1]
for i in range(1, len(prev_row)):
new_row.append(prev_row[i - 1] + prev_row[i])
new_row.append(1)
triangle.append(new_row)
return triangle
n = int(input())
triangle = pascal(n)
for row in triangle:
print(row)Contributors
lucashsu95