<Codesignal> Phone Call

Easy

Some phone usage rate may be described as follows:

• first minute of a call costs min1 cents,
• each minute from the 2nd up to 10th (inclusive) costs min2_10 cents
• each minute after 10th costs min11 cents.

You have s cents on your account before the call. What is the duration of the longest call (in minutes rounded down to the nearest integer) you can have?

Example

For min1 = 3, min2_10 = 1, min11 = 2, and s = 20, the output should be
phoneCall(min1, min2_10, min11, s) = 14.

Here's why:

• the first minute costs 3 cents, which leaves you with 20 - 3 = 17 cents;
• the total cost of minutes 2 through 10 is 1 * 9 = 9, so you can talk 9 more minutes and still have 17 - 9 = 8 cents;
• each next minute costs 2 cents, which means that you can talk 8 / 2 = 4 more minutes.

Thus, the longest call you can make is 1 + 9 + 4 = 14 minutes long.

Input/Output

• [execution time limit] 0.5 seconds (c)

• [input] integer min1

  Guaranteed constraints:
  1 ≤ min1 ≤ 10.

• [input] integer min2_10

  Guaranteed constraints:
  1 ≤ min2_10 ≤ 10.

• [input] integer min11

  Guaranteed constraints:
  1 ≤ min11 ≤ 10.

• [input] integer s

  Guaranteed constraints:
  2 ≤ s ≤ 500.

• [output] integer

[C] Syntax Tips

// Prints help message to the console
// Returns a string
char * helloWorld(char * name) {
    char * answer = malloc(strlen(name) + 8);
    printf("This prints to the console when you Run Tests");
    strcpy(answer, "Hello, ");
    strcat(answer, name);
    return answer;
}

---

Solution

int phoneCall(int min1,int min2_10,int min11,int s)
{
	return s<min1?0:s<min1+9*min2_10?(s-min1)/min2_10+1:(s-min1-9*min2_10)/min11+10;
}