## The challenge

A number m of the form 10x + y is divisible by 7 if and only if x − 2y is divisible by 7. In other words, subtract twice the last digit from the number formed by the remaining digits. Continue to do this until a number known to be divisible by 7 is obtained; you can stop when this number has *at most* 2 digits because you are supposed to know if a number of at most 2 digits is divisible by 7 or not.

The original number is divisible by 7 if and only if the last number obtained using this procedure is divisible by 7.

**Examples:**

1 – `m = 371 -> 37 − (2×1) -> 37 − 2 = 35`

; thus, since 35 is divisible by 7, 371 is divisible by 7.

The number of steps to get the result is `1`

.

2 – `m = 1603 -> 160 - (2 x 3) -> 154 -> 15 - 8 = 7`

and 7 is divisible by 7.

3 – `m = 372 -> 37 − (2×2) -> 37 − 4 = 33`

; thus, since 33 is not divisible by 7, 372 is not divisible by 7.

4 – `m = 477557101->47755708->4775554->477547->47740->4774->469->28`

and 28 is divisible by 7, so is 477557101. The number of steps is 7.

**Task:**

Your task is to return to the function `seven(m)`

(m integer >= 0) an array (or a pair, depending on the language) of numbers, the first being the *last* number `m`

with at most 2 digits obtained by your function (this last `m`

will be divisible or not by 7), the second one being the number of steps to get the result.

**Forth Note:**

Return on the stack `number-of-steps, last-number-m-with-at-most-2-digits`

**Examples:**

```
seven(371) should return [35, 1]
seven(1603) should return [7, 2]
seven(477557101) should return [28, 7]
```

## The solution in C

Option 1:

```
#include <stdlib.h>
int* seven (long long m) {
int* res = (int*)malloc (2 * sizeof (int));
int step = 0;
while (m > 99) {
m = m / 10 - 2 * (m % 10);
++step;
}
res[0] = m;
res[1] = step;
return res;
}
```

Option 2:

```
int* seven(long long m) {
static int res[2] = {};
int cnt = 0;
while (m > 99) {
long long a0 = m % 10;
m = (m - a0) / 10 - 2 * a0;
cnt++;
}
res[0] = (int) m;
res[1] = cnt;
return res;
}
```

Option 3:

```
#include <assert.h>
#include <stdlib.h>
int* seven(long long n) {
assert(n >= 0);
int *result = (int *) malloc(2 * sizeof(int));
if (! result)
return NULL;
int steps = 0;
for (; n > 99; ++steps)
n = n / 10 - n % 10 * 2;
result[0] = n;
result[1] = steps;
return result;
}
```

## Test cases to validate our solution

```
#include <criterion/criterion.h>
extern int *seven (long long m);
static void do_test (long long m, const int expected[2])
{
int *actual = seven(m);
cr_assert_arr_eq(actual, expected, 2 * sizeof *expected,
"for m = %lld, expected {%d, %d} but got {%d, %d}",
m, expected[0], expected[1], actual[0], actual[1]
);
}
#define ARRAY (const int[2])
Test(tests_suite, fixed_tests)
{
do_test(1021, ARRAY{10, 2});
do_test(477557101, ARRAY{28, 7});
do_test(477557102, ARRAY{47, 7});
do_test(1603, ARRAY{7, 2});
do_test(371, ARRAY{35, 1});
do_test(1369851, ARRAY{0, 5});
do_test(483, ARRAY{42, 1});
do_test(483595, ARRAY{28, 4});
do_test(0, ARRAY{0, 0});
do_test(170232833617690725ll, ARRAY {0, 16});
}
```