Midterm Question 2 - Makeup

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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: 36751 - Available when: 2017-11-15 14:10 - Due to: Unlimited | 4k views

26 Answers

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Hidden content!
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answered by (-108 points)
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prog.c: In function 'main':
prog.c:8:7: error: 'odd' undeclared (first use in this function)
     a=odd number;
       ^~~
prog.c:8:7: note: each undeclared identifier is reported only once for each function it appears in
prog.c:8:11: error: expected ';' before 'number'
     a=odd number;
           ^~~~~~
prog.c:9:7: error: 'not' undeclared (first use in this function)
     b=not odd number;
       ^~~
prog.c:9:11: error: expected ';' before 'odd'
     b=not odd number;
           ^~~
prog.c:10:12: error: lvalue required as left operand of assignment
     a*1+b*2=sum;
            ^
prog.c:11:21: error: 'last' undeclared (first use in this function)
     if (10-(sum%10)=last number)
                     ^~~~
prog.c:11:26: error: expected ')' before 'number'
     if (10-(sum%10)=last number)
                          ^~~~~~
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