UPC 代码问题

Question

我一直在努力找出为什么我的输出与预期不符。给出的样本是

Enter a UPC (enter a blank line to quit): 036000291453 Check digit should be: 2 Check digit is: 3 UPC is not valid

Enter a UPC (enter a blank line to quit): 036000291452 Check digit should be: 2 Check digit is: 2 UPC is valid

Enter a UPC (enter a blank line to quit): 014633149077 Check digit should be: 4 Check digit is: 7 UPC is not valid

Enter a UPC (enter a blank line to quit): 014633149074 Check digit should be: 4 Check digit is: 4 UPC is valid

Enter a UPC (enter a blank line to quit): 0853911765722 ERROR! UPC MUST have exactly 12 digits

Enter a UPC (enter a blank line to quit): 085391176572 Check digit should be: 2 Check digit is: 2 UPC is valid

Enter a UPC (enter a blank line to quit): Goodbye!

得到那个输出的算法是这样的从左到右，将奇数位置的数字相加（从1开始计数），并将结果乘以3。从左到右，将偶数位置的数字添加到步骤 1 中计算的总数取步骤 2 的结果并计算除以 10 的余数（结果模 10）。如果余数不为零，用 10 减去这个余数得到校验位。如果余数为零，则校验位应为0。

    String str1 = validinput(in);
    int odd1 = odd(str1);
    int even1 = even(str1);
    int f = (odd1+even1)%10;
    if(f != 0){
        f = 10-f;
    }
    System.out.println(odd1);
    System.out.println(even1);

    System.out.println("Check digit should be: "+f);
    System.out.println("Check digit is: "+str1.charAt(11));
    int y = Character.getNumericValue(str1.charAt(11));
    if (f == y){
        System.out.println("UPC is valid");
    }
    else{
        System.out.println("UPC is not valid");
    }
}


private static String validinput(Scanner inScanner){

    System.out.print("Enter a UPC (enter a blank line to quit): ");
    String str = inScanner.nextLine();


    while(str.length() != 12){
        if (str.length() == 0){
            System.out.println("Goodbye");
            break;
        }
        else{
        System.out.println("ERROR! UPC MUST have exactly 12 digits");

        System.out.print("Enter a UPC (enter a blank line to quit): ");
        str = inScanner.nextLine();
        }
    }
    return str;
}


private static int odd(String input){
    int i = 1;
    char ch;
    int sumOdd = 0;
    while (i < 11){
        ch = input.charAt(i);
        int x = Character.getNumericValue(ch);
        sumOdd = x +sumOdd;
        i += 2;
    }
    int Mx3=sumOdd*3;
    return Mx3;
}


private static int even(String input){
    int i = 0;
    char ch;
    int sumEven = 0;
    while (i < 11){
        ch = input.charAt(i);
        int x = Character.getNumericValue(ch);
        sumEven = x +sumEven;
        i += 2;
    }

    return sumEven;
 }

Answer 1

charAt() 使用基于 null 的索引，但指令希望您使用基于 1 的索引。因此，在 odd() 中以 i = 0 开始。在 even() 中以 i = 1 开始。

其次，您使用Character.getNumericValue(ch)获取字符的unicode代码点值，但说明要求您使用数字值。所以，请改用 Integer.parseInt(ch.ToString())。

private static int odd(String input){
    int i = 0;
    char ch;
    int sumOdd = 0;
    while (i < 11){
        ch = input.charAt(i);
        int x = Integer.parseInt(ch.ToString());
        sumOdd = x +sumOdd;
        i += 2;
    }
    int Mx3=sumOdd*3;
    return Mx3;
}


private static int even(String input){
    int i = 1;
    char ch;
    int sumEven = 0;
    while (i < 11){
        ch = input.charAt(i);
        int x = Integer.parseInt(ch.ToString());
        sumEven = x +sumEven;
        i += 2;
    }

    return sumEven;
 }

代码未测试

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