Valid Card Number

Question

Я прошел обычный тест и рандомный, но не могу пройти advanced_tests

Условие - In this Kata, you will implement the Luhn Algorithm, which is used to help validate credit card numbers.

Given a positive integer of up to 16 digits, return true if it is a valid credit card number, and false if it is not.

Here is the algorithm:

Double every other digit, scanning from right to left, starting from the second digit (from the right).

Another way to think about it is: if there are an even number of digits, double every other digit starting with the first; if there are an odd number of digits, double every other digit starting with the second:

1714 ==> [1*, 7, 1*, 4] ==> [2, 7, 2, 4]

12345 ==> [1, 2*, 3, 4*, 5] ==> [1, 4, 3, 8, 5]

891 ==> [8, 9*, 1] ==> [8, 18, 1] If a resulting number is greater than 9, replace it with the sum of its own digits (which is the same as subtracting 9 from it):

[8, 18*, 1] ==> [8, (1+8), 1] ==> [8, 9, 1]

or:

[8, 18*, 1] ==> [8, (18-9), 1] ==> [8, 9, 1] Sum all of the final digits:

[8, 9, 1] ==> 8 + 9 + 1 = 18 Finally, take that sum and divide it by 10. If the remainder equals zero, the original credit card number is valid.

#include <vector>
#include <algorithm>
using namespace std;

class Kata {
  public:
  static bool validate(long long int n) {
    long long  a, b=0;
        std::vector<int>arr;
        while (n > 0)
        {
            a = n % 10;
            b = b * 10 + a;
            n = n / 10;
        }
        int z = 0;
        while (b > 0)
        {
            z = b % 10;
            arr.push_back(z);
            b = b / 10;
        }


    int sum = 0;

   
    if (arr.size() % 2 == 0) {
        for (unsigned long i = 0; i < arr.size(); i = i + 2) {

            arr[i] = arr[i] + arr[i];
            cout << arr[i] << " ";
        }
        cout << "\n";

        for (unsigned long i = 0; i < arr.size(); ++i) {
            cout << arr[i] << " ";
        }
        cout << "\n";

        for (unsigned long i = 0; i < arr.size(); i = i + 2) {
            if (arr[i] > 10) {
                arr[i] = arr[i] - 9;
            }

            cout << arr[i] << " ";

        }

        for (unsigned long i = 0; i < arr.size(); i++) {
            sum = sum + arr[i];
        }
        cout << sum << endl;




    }else{
        for (unsigned long i = 1; i < arr.size(); i = i + 2) {

            arr[i] = arr[i] + arr[i];
            cout << arr[i] << " ";
        }
        cout << "\n";

        for (unsigned long i = 0; i < arr.size(); ++i) {
            cout << arr[i] << " ";
        }
        cout << "\n";

        for (unsigned long i = 0; i < arr.size(); i = i + 1) {
            if (arr[i] >=10){
                arr[i] = arr[i] - 9;
            }

            cout << arr[i] << " ";

        }

        for (unsigned long i = 0; i < arr.size(); i++) {
            sum = sum + arr[i];
        }
        cout<<" "<<endl;
        cout << sum << endl;

    }


    
    if (sum % 10 == 0) {
       return true;
    } else {
       return false;
    }

        
    }
  };

Answer 1

Вот так попробуйте:

bool validate(long long int n){
    std::string input = std::to_string(n);
    int len = input.length(),
        number = 0,
        sum = 0;
    bool len_ = len%2

    for(int i=len; i>0; --i){
        number = input[i]-'0';
        if((len_==1 && i%2==0) || (len_==0 && i%2==1)) {
            number *= 2;
            if(number > 9)
                number -= 9;
        }
        sum += number;
    }
    return !(sum%10);
}

Answer 2

Воистину, зачем просто, если можно сложно...

static bool validate(long long int n) {
    int s = 0;
    for(int dbl = 2; n; n/= 10)
    {
        int m = (n%10)*(dbl = 3-dbl);
        if (m > 9) m -= 9;
        s += m;
    }
    return s%10==0;
}

P.S. В следующий раз в таких случаях давайте URL проверялки, а то самому искать - лишние затраты времени, а без него - сплошной испорченный телефон :)

БЛОГ НА HUSL

Valid Card Number

Ответы (2 шт):