Description: Our football team finished the championship. The result of each match look like "x:y". Results of all matches are recorded in the collection.

For example: ["3:1", "2:2", "0:1", ...]

Write a function that takes such collection and counts the points of our team in the championship. Rules for counting points for each match:

if x>y - 3 points if x<y - 0 point if x=y - 1 point Notes:

there are 10 matches in the championship 0 <= x <= 4 0 <= y <= 4

A)

```c int points(const char* const games[10]) { int i = 0, point = 0;; while (i < 10) { if (games[i][0] > games[i][2]) point += 3; else if (games[i][0] < games[i][2]) point += 0; else if (games[i][0] == games[i][2]) point += 1; i++; } return point; }

another solution #include <stdlib.h>

int points(char* games[]) { int p = 0; for (int i = 0; i < 10; ++i) { int x = games[i][0] - '0'; int y = games[i][2] - '0'; p += (x > y) * 3 + (x == y); } return p; }

another solution #include <stdlib.h>

int points(char* games[]) {

int points = 0; 
int x, y;

for (int i = 0; i < 10; i++)
{
  sscanf(games[i], "%d:%d", &x, &y);

  if (x > y) points += 3;
  else if (x == y) points += 1;
}

return points;

} -> sscanf() 로 받아서 x, y에 저장할 수 있는 듯. 원리는 배열에 들어갈때 input stream을 열어주는 듯?

Description: Sum all the numbers of the array (in F# and Haskell you get a list) except the highest and the lowest element (the value, not the index!). (The highest/lowest element is respectively only one element at each edge, even if there are more than one with the same value!)

Example:

{ 6, 2, 1, 8, 10 } => 16 { 1, 1, 11, 2, 3 } => 6

If array is empty, null or None, or if only 1 Element exists, return 0. Note:In C++ instead null an empty vector is used. In C there is no null. ;-)

-- There's no null in Haskell, therefore Maybe [Int] is used. Nothing represents null. Have fun coding it and please don't forget to vote and rank this kata! :-)

I have created other katas. Have a look if you like coding and challenges.

A)

```c int sum(int* numbers, int numbersCount) { int mid_sum = 0; int min_idx = 0; int max_idx = 0; for (int i = 1; i < numbersCount; i++) { if (numbers[min_idx] > numbers[i]) min_idx = i; } for (int i = 1; i < numbersCount; i++) { if (numbers[max_idx] < numbers[i]) max_idx = i; } for (int i = 0; i < numbersCount; i++) { if (i != min_idx & i != max_idx) mid_sum += numbers[i]; }
return mid_sum; }

another solution 남들은 for 문 하나로 모든 원소 sum 하면서 max랑 min 찾아내고 나중에 max, min을 sum에서 뻄.

Description: Given a string of digits, you should replace any digit below 5 with '0' and any digit 5 and above with '1'. Return the resulting string.

A)

```c #include <string.h>

void fakeBin(const char *digits, char *buffer) { // for (size_t i = 0; i < strlen(digits); i++) // { // if (digits[i] >= '0' && digits[i] < '5') // buffer[i] = '0'; // else if (digits[i] >= '5' && digits[i] <= '9') // buffer[i] = '1'; // } // buffer[strlen(buffer)] = '\0'; size_t len = strlen(digits); for (size_t i = 0; i < len; i++) buffer[i] = (digits[i] >= '5' ? '1' : '0'); buffer[len] = '\0'; } -> 주석코드는 왜 안됐지? -> buffer len을 재니까 안됐지 ㅋㅋ

Description: Clock shows h hours, m minutes and s seconds after midnight.

Your task is to write a function which returns the time since midnight in milliseconds.

Example: h = 0 m = 1 s = 1

result = 61000 Input constraints:

0 <= h <= 23 0 <= m <= 59 0 <= s <= 59

A)

```c int past(int h, int m, int s) { int result; return result = h * 3600000 + m * 60000 + s * 1000; }

Description: Note: This kata is inspired by Convert a Number to a String!. Try that one too.

Description We need a function that can transform a string into a number. What ways of achieving this do you know?

Note: Don't worry, all inputs will be strings, and every string is a perfectly valid representation of an integral number.

Examples "1234" --> 1234 "605" --> 605 "1405" --> 1405 "-7" --> -7

A)

```c int string_to_number(const char *src) { return atoi(src); }

Description: Description: Make a simple function called greet that returns the most-famous "hello world!".

Style Points Sure, this is about as easy as it gets. But how clever can you be to create the most creative hello world you can think of? What is a "hello world" solution you would want to show your friends?

A)

```c //Write a function greet that returns "hello world!" char *greet(void) { char *ptr = "hello world!"; return ptr; }

another solution #include <string.h> #include <stdio.h> #include <stdlib.h>

char *greet() { char *code = "++++++++++[>++++++++++>+++++++++++>++++++++++++>+++<<<<-]>++++.---.>--..+++.>>++.<-.<.>-----.<---.<-.>>>+.";

char byteArr[] = {0x00, 0x00, 0x00, 0x00, 0x00}; int pointer = 0; int outPointer = 0; char out[14] = ""; int length = (int)strlen(code); for (int i = 0; i < length; i++) { char chr = code[i]; switch (chr) { case '>': pointer++; break;

case '<':
  pointer--;
  break;

case '+':
  byteArr[pointer]++;
  break;

case '-':
  byteArr[pointer]--;
  break;

case '.':
  out[outPointer] = byteArr[pointer];
  outPointer++;
  break;

case '[':
  if (byteArr[pointer] == 0x00)
    for (int j = 0; j < length; j++)
      if (code[j] == ']')
        i = j;
  break;

case ']':
  if (byteArr[pointer] != 0x00)
    for (int j = 0; j < length; j++)
      if (code[j] == '[')
        i = j;
  break;
}

}

char* result = malloc(strlen(out) + 1);

strcpy(result, out);

return result;

} -> ??

Description: It's the academic year's end, fateful moment of your school report. The averages must be calculated. All the students come to you and entreat you to calculate their average for them. Easy ! You just need to write a script.

Return the average of the given array rounded down to its nearest integer.

The array will never be empty.

A)

```c #include <stddef.h>

int get_average(const int *marks, size_t count) { int sum = 0; for (size_t i = 0; i < count; i++) sum += marks[i]; return sum / count; }

Description: Given a set of numbers, return the additive inverse of each. Each positive becomes negatives, and the negatives become positives.

invert([1,2,3,4,5]) == [-1,-2,-3,-4,-5] invert([1,-2,3,-4,5]) == [-1,2,-3,4,-5] invert([]) == [] Notes: All values are greater than INT_MIN The input should be modified, not returned.

A)

```c #include <stddef.h>

void invert(int *values, size_t values_size) { for (size_t i = 0; i < values_size; i++) values[i] *= -1; }

Description: Given an array of integers, return a new array with each value doubled.

For example:

[1, 2, 3] --> [2, 4, 6]

A)

```c #include <stddef.h>

// return a new, dynamically allocated array with each element doubled. int maps(const int *arr, size_t size) { int *tmp = (int)malloc(sizeof(int) * size); if (!tmp) return NULL; for (size_t i = 0; i < size; i++) tmp[i] = arr[i] * 2; return tmp; }

another solution int maps(const int *arr, size_t size) { int a = (int*)malloc(size * sizeof(int)); for (int i = 0; i < size; a[i++] = arr[i] * 2); return a; } -> 이런것도 되넹

Description: Given an array of integers.

Return an array, where the first element is the count of positives numbers and the second element is sum of negative numbers.

If the input array is empty or null, return an empty array.

Example For input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, -11, -12, -13, -14, -15], you should return [10, -65].

A)

```c #include <stddef.h>

void count_positives_sum_negatives( int *values, size_t count, int *positivesCount, int *negativesSum) { for (size_t i = 0; i < count; i++) { if (values[i] > 0) *positivesCount = *positivesCount + 1; else if (values[i] < 0) *negativesSum += values[i]; } } -> 0 is not positive

Description: Can you find the needle in the haystack?

Write a function findNeedle() that takes an array full of junk but containing one "needle"

After your function finds the needle it should return a message (as a string) that says:

"found the needle at position " plus the index it found the needle, so:

find_needle(['hay', 'junk', 'hay', 'hay', 'moreJunk', 'needle', 'randomJunk']) should return "found the needle at position 5"

A)

```c #include <stddef.h> #include <stdlib.h> #include <string.h>

char find_needle(const char *haystack, size_t count) { char *buf = calloc(100, sizeof(char)); char *tmp = malloc(256); if (!buf || !tmp) return NULL; size_t i = 0; strcpy(buf, "found the needle at position "); while (i < count) { if (strcmp(haystack[i], "needle") == 0) break; i++; } sprintf(tmp, "%d", i); strcat(buf, tmp); return buf; }

another solution #include <stdio.h>

char find_needle(const char **haystack, size_t count) { for(int i=0; i<count;++i) { if(!strcmp(haystack[i], "needle")) // strcmp will return 0 if true, so we need '!' to it to work { char buff; asprintf(&buff, "found the needle at position %d", i); return buff; } } } -> asprintf() .. i의 숫자를 넣은 문자열(read only)을 buff에 저장.

another solution #include <stddef.h> #include <stdlib.h> #include <string.h>

char find_needle(const char *haystack, size_t count) { while (strcmp(haystack[--count], "needle")); char *buf = malloc(128); sprintf(buf, "found the needle at position %d", count); return buf; } -> sprintf() .. 위와 비슷.

Description: Your task is to create a function that does four basic mathematical operations.

The function should take three arguments - operation(string/char), value1(number), value2(number). The function should return result of numbers after applying the chosen operation.

Examples basicOp('+', 4, 7) // Output: 11 basicOp('-', 15, 18) // Output: -3 basicOp('*', 5, 5) // Output: 25 basicOp('/', 49, 7) // Output: 7

A)

```c int basic_op(char op, int value1, int value2) { int cul = 0; if (op == '+') cul = value1 + value2; else if (op == '-') cul = value1 - value2; else if (op == '*') cul = value1 * value2; else if (op == '/') cul = value1 / value2; return cul; }

another solution int basic_op(char op, int x, int y) { return op == 43 ? x + y : op == 45 ? x - y : op == 42 ? x * y : x / y; } -> false 면 다음 기호로 넘어가는 방식. 숫자는 아스키코드로 연산기호들.

Description: Create a function that checks if a number n is divisible by two numbers x AND y. All inputs are positive, non-zero digits.

Examples:

1) n = 3, x = 1, y = 3 => true because 3 is divisible by 1 and 3 2) n = 12, x = 2, y = 6 => true because 12 is divisible by 2 and 6 3) n = 100, x = 5, y = 3 => false because 100 is not divisible by 3 4) n = 12, x = 7, y = 5 => false because 12 is neither divisible by 7 nor 5

A)

```c #include <stdbool.h>

bool isDivisible(int n, int x, int y) { return ((n % x == 0 && n % y == 0) ? true : false); }

Description: Nathan loves cycling.

Because Nathan knows it is important to stay hydrated, he drinks 0.5 litres of water per hour of cycling.

You get given the time in hours and you need to return the number of litres Nathan will drink, rounded to the smallest value.

For example:

time = 3 ----> litres = 1

time = 6.7---> litres = 3

time = 11.8--> litres = 5

A)

```c int Liters(double time) { int drink = 0; int save = (int)time; drink = save / 2; return drink; }

another solution int Liters(double time) { return (int)(time * 0.5); }

Description: Complete the square sum function so that it squares each number passed into it and then sums the results together.

For example, for [1, 2, 2] it should return 9 because 1^2 + 2^2 + 2^2 = 9.

A)

```c #include <stddef.h>

int square_sum(const int *values, size_t count) { int sum = 0; for (; count; count--) sum += values[count - 1] * values[count - 1]; return sum; }

Description: Consider an array/list of sheep where some sheep may be missing from their place. We need a function that counts the number of sheep present in the array (true means present).

For example,

{ true, true, true, false, true, true, true, true, true, false, true, false, true, false, false, true, true, true, true, true, false, false, true, true } The correct answer would be 17.

Hint: Don't forget to check for bad values like null/undefined

A.

```c #include <stdbool.h> #include <stddef.h>

size_t count_sheep(const bool *sheep, size_t count) { size_t find = 0; while (count--) { if (sheep[count]) find++; } return find; }

another solution size_t count_sheep(const bool *sheep, size_t count) { int n=0; while(count--) n += sheep[count]; return n; } -> true는 1, false는 0이기 때문에 가능한 식.

Description: We need a function that can transform a number into a string.

What ways of achieving this do you know?

In C, return a dynamically allocated string that will be freed by the test suite.

Examples: 123 --> "123" 999 --> "999"

A)

```c #include <stdlib.h>

const char* number_to_string(int number) { char *ptr; int len = 0; long long nb = (long long)number;

if (number < 0) nb *= -1;

while (nb != 0) { nb = nb / 10; len++; }

if (number > 0) ptr = (char)malloc(sizeof(char) * len + 1); else { len++; ptr = (char)malloc(sizeof(char) * len + 1); }

if (!ptr) return NULL;

ptr[len] = '\0';

nb = (long long)number;

if (number < 0) nb *= -1;

while (len--) { ptr[len] = '0' + (nb % 10); nb /= 10; }

if (number < 0) ptr[0] = '-';

return ptr; }

another solution #include <stdio.h>

const char* number_to_string(int number) { char *s; asprintf(&s, "%d", number); return s; }

another solution #include <stdlib.h>

const char* number_to_string(int number) { char *itoa = malloc(256); sprintf(itoa, "%d", number); return itoa; } -> sprintf() function으로 첫 매개변수에 받는 버퍼에 number를 문자열로 저장하는 듯.

another solution #include <stdlib.h>

const char* number_to_string(int number) {

size_t bufferLength = snprintf(NULL, 0, "%d", number); //Find the length of the string conversion
char* buffer = malloc(bufferLength + 1); //dynamically allocate the string

snprintf(buffer, bufferLength + 1, "%d", number); //populate the string

return buffer; //return the string

}

Description: Given an array of integers your solution should find the smallest integer.

For example:

Given [34, 15, 88, 2] your solution will return 2 Given [34, -345, -1, 100] your solution will return -345 You can assume, for the purpose of this kata, that the supplied array will not be empty.

A)

```c #include <stddef.h>

int find_smallest_int(int *vec, size_t len) { size_t min_idx = 0; size_t i = 1; while (i < len) { if (vec[min_idx] > vec[i]) min_idx = i; i++; } return vec[min_idx]; }

another solution 사망연산자 쓰기

Description: Simple, remove the spaces from the string, then return the resultant string.

For C, you must return a new dynamically allocated string.

A)

```c #include <stdlib.h>

char no_space(const char *str_in) { char *ptr; int i = 0, j = 0, len = 0; while (str_in[i]) { if (str_in[i] != ' ') len++; i++; } if (!(ptr = (char)malloc(sizeof(char) * len + 1))) return NULL; i = 0; while (str_in[i]) { if (str_in[i] != ' ') ptr[j++] = str_in[i]; i++; } ptr[len] = '\0'; return ptr; }

another solution char no_space(char *s) { char *res = strdup(s), *q = res; for (; *s; s++) if (s != ' ') *q++ = *s; return *q = 0, res; } -> strdup() function : 문자열 동적할당해주고 카피해주는데 이렇게 하면 직접 쓸 메모리공간보다 더 할당해주기 때문에 낭비될 수 있음.

Description: Summation Write a program that finds the summation of every number from 1 to num. The number will always be a positive integer greater than 0.

For example:

summation(2) -> 3 1 + 2

summation(8) -> 36 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8

A)

```c int summation(int num) { int sum = 0; for (int i = 1; i <= num; i++) sum += i; return sum; }

another solution int summation(int num) { return num * (num + 1) / 2; } -> ex: num = 10 / (10 * 11) / 2 = 55 / 가우스 공식.