CS50信用问题万事达卡号为什么要无效

Question

我正在尝试为 CS50 编写信用问题代码，它适用于大多数号码，但是当我运行支票时，我收到一条错误消息，提示我的代码正在返回 Mastercard 5555 5555 5555 4444，应该是无效的。手动弄清楚，我不明白为什么那张卡会无效。每第二个数字加倍：10 (1) + 10 (1) + 10 (1) + 10 (1) + 10 (1) + 10 (1) + 8 + 8 = 22。然后 22 + 5 + 5 + 5 + 5 + 5 + 5 + 4 + 4 = 50。50 是 10 的倍数，数字长 16 位，以 55 开头，那为什么不是万事达卡？

我对 post 我的代码感到尴尬，因为我知道它非常麻烦，而且肯定有更快、更优雅的方法来做同样的事情，但这是我第一次尝试编程，所以我有以我脑海中唯一合乎逻辑的方式去做——而且我只知道 class.

中显示的脚本

总之，就到这里了，大家不要笑，哈哈！

// To determine if a user's credit card number is valid and which company it belongs to

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

int main(void)
{
// To get user input of credit card number
    long card;
    do
    {
        card = get_long("What is your credit card number: ");
    }
    while (card < 0);
// To divide number into digits
    int a = card % 10;
    int b = card % 100 - a; 
    int bb = b / 10;
    int c = card % 1000 - b - a; 
    int cc = c / 100;
    int d = card % 10000 - c - b - a; 
    int dd = d / 1000;
    int e = card % 100000 - d - c - b - a; 
    int ee = e / 10000;
    int f = card % 1000000 - e - d - c - b - a; 
    int ff = f / 100000;
    int g = card % 10000000 - f - e - d - c - b - a; 
    int gg = g / 1000000;
    int h = card % 100000000 - g - f - e - d - c - b - a; 
    int hh = h / 10000000;
    int i = card % 1000000000 - h - g - f - e - d - c - b - a; 
    int ii = i / 100000000;
    int j = card % 10000000000 - i - h - g - f - e - d - c - b - a; 
    int jj = j / 1000000000;
    int k = card % 100000000000 - j - i - h - g - f - e - d - c - b - a; 
    int kk = k / 10000000000;
    int l = card % 1000000000000 - k - j - i - h - g - f - e - d - c - b - a; 
    int ll = l / 100000000000;
    int m = card % 10000000000000 - l - k - j - i - h - g - f - e - d - c - b - a; 
    int mm = m / 1000000000000;
    int n = card % 100000000000000 - m - l - k - j - i - h - g - f - e - d - c - b - a; 
    int nn = n / 10000000000000;
    int o = card % 1000000000000000 - n - m - l - k - j - i - h - g - f - e - d - c - b - a; 
    int oo = o / 100000000000000;
    int p = card % 10000000000000000 - o - n - m - l - k - j - i - h - g - f - e - d - c - b - a; 
    int pp = p / 1000000000000000;

// To multiply each digit and reduce to one digit
    int aaa = a * 2;
    if (aaa > 9)
    {
        int ab = aaa % 10; 
        int ac = aaa % 100 - ab; 
        int ad = ac / 10;
        aaa = ad + ab;
    }
    int bbb = bb * 2;
    if (bbb > 9)
    {
        int ba = bbb % 10; 
        int bc = bbb % 100 - ba; 
        int bd = bc / 10;
        bbb = bd + ba;
    }
    int ccc = cc * 2;
    if (ccc > 9)
    {
        int cb = ccc % 10; 
        int ca = ccc % 100 - cb; 
        int cd = ca / 10;
        ccc = cd + cb;
    }
    int ddd = dd * 2;
    if (ddd > 9)
    {
        int db = ddd % 10; 
        int dc = ddd % 100 - db; 
        int da = dc / 10;
        ddd = da + db;
    }
    int eee = ee * 2;
    if (eee > 9)
    {
        int eb = eee % 10;
        int ec = eee % 100 - eb; 
        int ed = ec / 10;
        eee = ed + eb;
    }
    int fff = ff * 2;
    if (fff > 9)
    {
        int fb = fff % 10; 
        int fc = fff % 100 - fb; 
        int fd = fc / 10;
        fff = fd + fb;
    }
    int ggg = gg * 2;
    if (ggg > 9)
    {
        int gb = ggg % 10; 
        int gc = ggg % 100 - gb; 
        int gd = gc / 10;
        ggg = gd + gb;
    }
    int hhh = hh * 2;
    if (hhh > 9)
    {
        int hb = hhh % 10; 
        int hc = hhh % 100 - hb; 
        int hd = hc / 10;
        hhh = hd + hb;
    }
    int iii = ii * 2;
    if (iii > 9)
    {
        int ib = iii % 10; 
        int ic = iii % 100 - ib; 
        int id = ic / 10;
        iii = id + ib;
    }
    int jjj = jj * 2;
    if (jjj > 9)
    {
        int jb = jjj % 10; 
        int jc = jjj % 100 - jb; 
        int jd = jc / 10;
        jjj = jd + jb;
    }
    int kkk = kk * 2;
    if (kkk > 9)
    {
        int kb = kkk % 10; 
        int kc = kkk % 100 - kb; 
        int kd = kc / 10;
        kkk = kd + kb;
    }
    int lll = ll * 2;
    if (lll > 9)
    {
        int lb = lll % 10; 
        int lc = lll % 100 - lb; 
        int ld = lc / 10;
        lll = ld + lb;
    }
    int mmm = mm * 2;
    if (mmm > 9)
    {
        int mb = mmm % 10; 
        int mc = mmm % 100 - mb; 
        int md = mc / 10;
        mmm = md + mb;
    }
    int nnn = nn * 2;
    if (nnn > 9)
    {
        int nb = nnn % 10; 
        int nc = nnn % 100 - nb; 
        int nd = nc / 10;
        nnn = nd + nb;
    }
    int ooo = oo * 2;
    if (ooo > 9)
    {
        int ob = ooo % 10; 
        int oc = ooo % 100 - ob; 
        int od = oc / 10;
        ooo = od + ob;
    }
    int ppp = pp * 2;
    if (ppp > 9)
    {
        int pb = ppp % 10; 
        int pc = ppp % 100 - pb; 
        int pd = pc / 10;
        ppp = pd + pb;
    }
// To determine if card is valid

    int pra = bbb + ddd + fff + hhh + jjj + lll + nnn + ppp;
    int prb = pra + aaa + ccc + eee + ggg + iii + kkk + mmm + ooo;
// To determine if card is AMEX
    if (card >= 370000000000000 && card < 380000000000000)
    {
        printf("AMEX\n");
    }
    else if (card >= 340000000000000 && card < 350000000000000)
    {
       printf("AMEX\n");
    }
// To determine if card is MASTERCARD
    else if (card >= 5100000000000000 && card < 5500000000000000)
    {
       printf("MASTERCARD\n");
    }
// To determine if card is VISA
    else if (card >= 4000000000000 && card < 5000000000000)
    {
       printf("VISA\n");
    }
    else if (card >= 4000000000000000 && card < 5000000000000000)
    {
       printf("VISA\n");
    }
// To determine if card is INVALID
    else
    {
        printf("INVALID\n");
    }
}

Answer 1

我看到您经常使用单个整数，代码变得超长且不可维护。这是一个非常糟糕的设计。

而不是整数，使用string来存储信用卡号。要达到信用卡号中的个人号码，您可以使用

card = get_string("What is your credit card number: ");

card[i]

i是信用卡号的索引。从零开始，循环遍历所有数字，然后进行计算。

CS50信用问题万事达卡号为什么要无效

CS50 credit problem why should mastercard number be invalid

c

credit-card

mastercard

cs50