Comments (2)
#3주차 숙제 시저암호 소스코드
def solution(s, n):
a = ''
for i in range(len(s)):
if ord(s[i]) == 32: #s[i]가 공백일때
a = a + s[i]
elif ord(s[i]) + n > 122: #s[i] + n이 z를 넘어갈때
a = a + chr(ord(s[i]) + n - 26)
elif ord(s[i]) <= 90 and ord(s[i]) + n > 90: #s[i] + n이 Z를 넘어갈때
a = a + chr(ord(s[i]) + n - 26)
else: #s[i] + n이 z나 Z를 넘어가지 않고 공백도 아닐때
a = a + chr(ord(s[i]) + n)
return a
#3주차 숙제 알파벳 찾기
s = input()
for i in range(26):
check = 0
for k in range(len(s)):
if check == 0: # 알파벳이 중복되는 것을 방지
if ord(s[k]) == 97 + i:
print(k, end=" ")
check = 1
if check == 0: # 알파벳을 찾았으면 -1을 출력 안함
print("-1", end = " ")
from 2021-spring-python-tutoring.
과제 2번 시저암호 알파벳 찾기
def solution(s, n):
s = list(s) # 문자를 리스트 형식으로
for i in range(len(s)):
if s[i].isupper(): # 대문자
s[i]=chr((ord(s[i])-ord('A')+ n)%26+ord('A')) #%26을 통해 n을 0~25 유지
elif s[i].islower(): #소문자
s[i]=chr((ord(s[i])-ord('a')+ n)%26+ord('a'))
answer = "".join(s) #리스트에 합치기
return answer
#==================================================================================
A = input() #입력할 값
L = list(range(ord("a"), 123))
# print(ord("a")) #97
# print(ord('z')) #122
for i in L:
print(A.find(chr(i)),end= " ")
from 2021-spring-python-tutoring.
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from 2021-spring-python-tutoring.