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The 13-digit International Standard Book Number (ISBN) is a unique code that identifies a book commercially. The last digit is a check digit used for error detection. To calculate its value, each digit of the first twelve digits is alternately multiplied, from left to right, by 1 or 3. The products are summed up and divided by 10. The check digit is the remainder of the division subtracted from 10. If it is 10, it becomes 0. 

ISBN檢查號的計算方法: 前12個數字,奇數項 乘以 1 , 偶數項乘以 3

For example, assume that the first twelve digits are 978960931961.  

假設號碼前12位是978960931961

(a) (9*1 + 7*3 + 8*1 + 9*3 + 6*1 + 0*3 + 9*1 + 3*3 + 1*1 + 9*3 + 6*1 + 1*3) = 126 

(b) The check digit = 10 – (126 % 10) = 10 – 6 = 4  

檢查號 = 10 - 餘數 = 10 – (126 % 10) = 10 – 6 = 4  

So the valid ISBN is: 9789609319614

Write a program that reads a 13-digit ISBN and checks the last digit to see if it is valid or not. 

寫一個輸入13位ISBN號碼 判斷最後一個數字是否檢查碼即可

#Try 1:

Input:

9789609319614

Output: 

Valid ISBN

 

#Try 2:

Input:

9781482214505

Output: 

Valid ISBN

 

#Try 3:

Input:

9780132936493

Output: 

Invalid ISBN

 

[Exam] asked in Midterm by (12.1k points)
ID: 32706 - Available when: 2017-11-15 14:10 - Due to: Unlimited
| 14.2k views

49 Answers

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answered by (-286 points)
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Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
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answered by (-229 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
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answered by (-304 points)
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Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
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answered by (-249 points)
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Case 2: Correct output
Case 3: Correct output
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answered by (-284 points)
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35/100 answered by (-285 points)
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prog.c: In function 'main':
prog.c:12:11: warning: zero width in gnu_scanf format [-Wformat=]
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           ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
prog.c:12:11: warning: zero width in gnu_scanf format [-Wformat=]
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prog.c:12:11: warning: zero width in gnu_scanf format [-Wformat=]
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answered by (-323 points)
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answered by (-127 points)
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Case 2: Correct output
Case 3: Correct output
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