Split digits in array [closed] - c

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I'm totally beginner and I have problem in C. So I have array:
A[5] = {14, 2, 7, 3, 2};
And I want make something like this:
A[6] = {1, 4, 2, 7, 3, 2};
From 14 to 1, 4. Any idea how to do that?
So I making the problem from Cs50. Here is a link: https://docs.cs50.net/2018/x/psets/1/credit/credit.html
And here is my code:
#include <stdio.h>
#include <cs50.h>
#include <string.h>
#include <math.h>
int main(void)
{
long long cc_number = 0;
do
{
printf("Number: ");
cc_number = get_long_long();
}
while(cc_number <= 0);
int num_of_digits = 0;
long long valid = cc_number;
while(valid > 0)
{
valid /= 10;
num_of_digits++;
}
if (num_of_digits != 13 && num_of_digits != 15 && num_of_digits != 16 ) //checking if number have more or less than 13,15,16 digits
{
printf("Number is invalid!\n");
}
long long k = 1; //create array for store each number from the card
int A[16], d, num = 0;
for(num = 0; num < 16; num++)
{
d = (cc_number/(1*k)) % 10;
A[num] = d;
k *=10;
}
///////////////////////////////////////////////////////////////////////
if (num_of_digits == 16)
{
for (int i = 0; i < 16; i = i + 2)
{
A[i] *= 2; // multiplay each second digit by 2
printf("this is %i\n", A[i]);
}
}
else if (num_of_digits == 15 || num_of_digits == 13 )
{
int sum = 0;
for (int i = 1; i < 15; i = i + 2)
{
int y = A[i];
A[i] *= 2; // multiplay each second digit by 2
if (A[i] > 9) // try to split digit from array
{
y = A[i] % 10;
A[i] /= 10;
}
sum += A[i];
printf("this is %i\n", A[i]);
}
printf("this is sum %i\n", sum);
}
///////////////////////////////////////////////////////////////////////////
}
And this is what i want to do:
For the sake of discussion, let’s first underline every other digit,
starting with the number’s second-to-last digit:
378282246310005
Okay, let’s multiply each of the underlined digits by 2:
7•2 + 2•2 + 2•2 + 4•2 + 3•2 + 0•2 + 0•2
That gives us:
14 + 4 + 4 + 8 + 6 + 0 + 0
Now let’s add those products' digits (i.e., not the products themselves) together:
1 + 4 + 4 + 4 + 8 + 6 + 0 + 0 = 27
Now let’s add that sum (27) to the sum of the digits that weren’t multiplied by 2:
27 + 3 + 8 + 8 + 2 + 6 + 1 + 0 + 5 = 60
Yup, the last digit in that sum (60) is a 0, so my card is legit!

Answer 1

Assuming you want to create an array of single digit integers from array of multi-digit integers
#include <stdio.h>
#define MAX 1024
void printarr(int *a, int n) { // function to print array
for(int i = 0; i < n; i++) {
printf("%d ", a[i]);
}
printf("\n");
}
int main() {
int a[5] = {14, 2, 7, 3, 2};
int b[MAX];
int k = 0;
printarr(a, 5);
char s[MAX]; // char buffer to store char array
for(int i = 0; i < 5; i++) {
sprintf(s, "%d", a[i]);// convert int to char array
int j = 0;
while(s[j]!='\0') { // for each digit, create a new integer
b[k++] = s[j++] - '0';
}
}
printarr(b, k);
return 0;
}
Output:
14 2 7 3 2
1 4 2 7 3 2
Else for specific case
#include <stdio.h>
#define MAX 1024
void printarr(int *a, int n) { // function to print array
for(int i = 0; i < n; i++) {
printf("%d ", a[i]);
}
printf("\n");
}
int main() {
int a[5] = {14, 2, 7, 3, 2};
int b[6];
printarr(a, 5);
b[0] = 1;
b[1] = 4;
for(int i = 2; i < 6; i++) {
b[i] = a[i - 1];
}
printarr(b, 6);
return 0;
}
Output:
14 2 7 3 2
1 4 2 7 3 2

Answer 2

Here is a solution that does not use strings.
First of all, we have to determine how long the new array will be. To this end, we determine how many digits each entry has. For n > 0 the number of digits is ⌊log10(n) + 1⌋.
After that, we extract the digits. 123 can be split into its digits by using integer division / and modulo % (remainder of integer division):
123 % 10 = 3 least significant digit
123 / 10 = 12
repeat
12 % 10 = 2 second least significant digit
12 / 10 = 1
repeat
1 % 10 = 1 third least significant digit
1 / 10 = 0
end
As you can see, the digits are extracted from the back, therefore we also fill the output array from the back.
#include <stdio.h>
#include <math.h>
int digitCount(int n) {
if (n)
return (int) log10(n) + 1;
return 1;
}
int main() {
int inputLength = 5;
int input[] = {14, 1, 9, 0, 5819};
int outputLength = 0;
for (int i = 0; i < inputLength; ++i)
outputLength += digitCount(input[i]);
int output[outputLength];
int o = outputLength;
for (int i = inputLength - 1; i >= 0; --i) {
int n = input[i];
do {
output[--o] = n % 10;
n /= 10;
} while (n);
}
while (o < outputLength) {
printf("%d ", output[o++]);
}
}
clang -lm file.c && ./a.out prints 1 4 1 9 0 5 8 1 9.