以下是luhn演算法(mod10 check)的步驟從右到左每隔兩位數。如果這個「翻倍」結果是兩位數，則將兩位數相加
得到乙個數字的數字。
現在將步驟1中的所有單個數字相加。
將信用卡號碼中從右到左的奇數字數相加。
將步驟2和步驟3的結果相加。
如果步驟4的結果可被10整除，則卡號有效；否則無效。
我的輸出應該是card number valid / invalid
3710293 invalid
5190990281925290 invalid
3716820019271998 valid
37168200192719989 invalid
8102966371298364 invalid
6823119834248189 valid
這是密碼。def checkseconddigits(num):
length = len(num)
sum = 0
for i in range(length-2,-1,-2):
number = eval(num[i])
number = number * 2
if number > 9:
strnumber = str(number)
number = eval(strnumber[0]) + eval(strnumber[1])
sum += number
return sum
def odd_digits(num):
length = len(num)
sumodd = 0
for i in range(length-1,-1,-2):
num += eval(num[i])
return sumodd
def c_length(num):
length = len(num)
if num >= 13 and num <= 16:
if num [0] == "4" or num [0] == "5" or num [0] == "6" or (num [0] == "3" and num [1] == "7"):
return true
else:
return false
def main():
filename = input("what is the name of your input file? ")
infile= open(filename,"r")
cc = (infile.readline().strip())
print(format("card number", "20s"), ("valid / invalid"))
print("------------------------------------")
while cc!= "exit":
even = checkseconddigits(num)
odd = odd_digits(num)
c_len = c_length(num)
tot = even + odd
if c_len == true and tot % 10 == 0:
print(format(cc, "20s"), format("valid", "20s"))
else:
print(format(cc, "20s"), format("invalid", "20s"))
num = (infile.readline().strip())
main()
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