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User Input for Cipher & Number of Rails #3

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User Input for Cipher & Number of Rails #3

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11 changes: 9 additions & 2 deletions py/decrypt.py
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Original file line number Diff line number Diff line change
@@ -1,7 +1,8 @@
# ==============================================================================
#
# Use:
# decrypt("HWe o!lordll", 4)
# Enter the cipher: HWe o!lordll
# Enter the number of rails: 4
# => "Hello World!"
#
# ==============================================================================
Expand Down Expand Up @@ -41,4 +42,10 @@ def decrypt(s,n):
if rail == n-1 or rail == 0:
var = -var

return r
return r

cipher_text=input("Enter the cipher: ")
rail_len=input("Enter the number of rails: ")
rail_len = int(rail_len)
plain_text = decrypt(cipher_text, rail_len)
print(plain_text)