Question

我正在处理PSET1（CS50）的“信用”任务。我可以使总数对于示例数字正确运行： 4003600000000014 ，但对其他任何数字均无效：

378282246310005作为AMEX
371449635398431作为AMEX
5555555555554444作为万事达卡
4111111111111111作为Visa

我也很难让它成功识别公司。我尝试在其中添加评论。这是link to the assignment。弄清楚我要完成的工作会更好。

#include <cs50.h>
#include <math.h>
#include <stdio.h>

int dubsum;
int singlesum;
int tot;
long secondspot;

int main(void)
{
    long number = get_long("Number: ");

    //finds every other spot starting at the back and doubling
    for (int i = 1; i <= 17; i = i + 2)
    {
        // front is the string from the starting place of desired number; back truncates behind
        long front = pow(10, i + 1);
        long back = pow(10, i);
        long backsecond = pow(10, i - 1);
        long spot = round(((number % front) - (number % back)) / back);

        //sanity check 1
//        printf("for i: %i,\ndigit to be doubled: %li\n", i, spot);
        //if we find a valid 2nd spot #, double it and add together its parts, then reset spot to 0
        for (int dub = 2 * spot; spot != 0 && i < 16; spot = 0)
        {
            //for double digit numbers after doubling
            if (dub >= 10)
            {
                int a = (dub % 100 - dub % 10) / 10;
                int b = dub % 10;
                dubsum = dubsum + a + b;
            }
            else
            {
                dubsum = dubsum + dub;
            }
        }

        // calculating the other numbers
        if (i == 1)
        {
            singlesum = number % back;
        }
        else
        {
            secondspot = round(((number % back) - (number % backsecond)) / backsecond);
            singlesum = singlesum + secondspot;
        }

        //sanity check on counters
        //printf("digit to be added: %li\nrunning tally for dub sum: %i\n running tally for singlesum: %i\n\n\n", secondspot, dubsum, singlesum);


    }

    tot = dubsum + singlesum;
    printf("%i", tot);

    //check on known numbers
    if (tot == 20)
    {
        //AmEx check: 15 digit # starting with 34/37
        if (round(number / pow(10, 13)) == 34 || round(number / pow(10, 13)) == 37)
        {
            printf("AmEx\n");
            printf("%f", round(number / pow(10, 13)));
        }
        //MasterCard check: 16 digt # starting with 51-55, inclusive
        if (round(number / pow(10, 14)) >= 51 && round(number / pow(10, 14)) <= 55)
        {
            printf("Mastercard\n");
        }
        //Visa Check: 13 || 16 digit # starting with 4
        if (round(number / pow(10, 15)) == 4 || round(number / pow(10, 12)) == 4)
        {
            printf("Visa\n");
        }
    }
    else
    {
        printf("INVALID\n");
    }

}

Answer 1

由于未使用else if语句和不正确的参数来检查总和，因此拒绝了数字。

//check on known numbers
if (tot % 10 == 0)
{
    //AmEx check: 15 digit # starting with 34/37
    if (trunc(number / pow(10, 13)) == 34 || trunc(number / pow(10, 13)) == 37)
    {
        printf("AMEX\n");
    }
    //MasterCard check: 16 digt # starting with 51-55, inclusive
    else if (trunc(number / pow(10, 14)) >= 51 && trunc(number / pow(10, 14)) <= 55)
    {
        printf("MASTERCARD\n");
    }
    //Visa Check: 13 || 16 digit # starting with 4
    else if (trunc(number / pow(10, 15)) == 4 || trunc(number / pow(10, 12)) == 4)
    {
        printf("VISA\n");
    }
    else if (tot % 10 == 0)
    {
        printf("INVALID\n");
    }

}
else
{
    printf("INVALID\n");
}

为什么我的CS50信用卡支票解决方案只能使用一个号码？

1 个答案: