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Easy

You are playing an RPG game. Currently your experience points (XP) total is equal to experience. To reach the next level your XP should be at least at threshold. If you kill the monster in front of you, you will gain more experience points in the amount of the reward.

Given values experience, threshold and reward, check if you reach the next level after killing the monster.

Example

• For experience = 10, threshold = 15, and reward = 5, the output should be
  reachNextLevel(experience, threshold, reward) = true;
• For experience = 10, threshold = 15, and reward = 4, the output should be
  reachNextLevel(experience, threshold, reward) = false.

Input/Output

• [execution time limit] 0.5 seconds (c)

• [input] integer experience

  Guaranteed constraints:
  3 ≤ experience ≤ 250.

• [input] integer threshold

  Guaranteed constraints:
  5 ≤ threshold ≤ 300.

• [input] integer reward

  Guaranteed constraints:
  2 ≤ reward ≤ 65.

• [output] boolean

  • true if you reach the next level, false otherwise.

[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;
}

bool reachNextLevel(int experience,int threshold,int reward)
{
	return experience+reward>=threshold;
}

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;
}

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;
}

Easy

One night you go for a ride on your motorcycle. At 00:00 you start your engine, and the built-in timer automatically begins counting the length of your ride, in minutes. Off you go to explore the neighborhood.

When you finally decide to head back, you realize there's a chance the bridges on your route home are up, leaving you stranded! Unfortunately, you don't have your watch on you and don't know what time it is. All you know thanks to the bike's timer is that n minutes have passed since 00:00.

Using the bike's timer, calculate the current time. Return an answer as the sum of digits that the digital timer in the format hh:mm would show.

Example

• For n = 240, the output should be
  lateRide(n) = 4.

  Since 240 minutes have passed, the current time is 04:00. The digits sum up to 0 + 4 + 0 + 0 = 4, which is the answer.

• For n = 808, the output should be
  lateRide(n) = 14.

  808 minutes mean that it's 13:28 now, so the answer should be 1 + 3 + 2 + 8 = 14.

Input/Output

• [execution time limit] 0.5 seconds (c)

• [input] integer n

  The duration of your ride, in minutes. It is guaranteed that you've been riding for less than a day (24 hours).

  Guaranteed constraints:
  0 ≤ n < 60 · 24.

• [output] integer

  • The sum of the digits the digital timer would show.

[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;
}

int lateRide(int n)
{
	return n/600+(n/60)%10+(n%60)/10+n%10;
}

Easy

Given a divisor and a bound, find the largest integer N such that:

• N is divisible by divisor.
• N is less than or equal to bound.
• N is greater than 0.

It is guaranteed that such a number exists.

Example

For divisor = 3 and bound = 10, the output should be
maxMultiple(divisor, bound) = 9.

The largest integer divisible by 3 and not larger than 10 is 9.

Input/Output

• [execution time limit] 0.5 seconds (c)

• [input] integer divisor

  Guaranteed constraints:
  2 ≤ divisor ≤ 10.

• [input] integer bound

  Guaranteed constraints:
  5 ≤ bound ≤ 100.

• [output] integer

  • The largest integer not greater than bound that is divisible by divisor.

[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;
}

int maxMultiple(int divisor,int bound)
{
	return bound-bound%divisor;
}

Easy

Your friend advised you to see a new performance in the most popular theater in the city. He knows a lot about art and his advice is usually good, but not this time: the performance turned out to be awfully dull. It's so bad you want to sneak out, which is quite simple, especially since the exit is located right behind your row to the left. All you need to do is climb over your seat and make your way to the exit.

The main problem is your shyness: you're afraid that you'll end up blocking the view (even if only for a couple of seconds) of all the people who sit behind you and in your column or the columns to your left. To gain some courage, you decide to calculate the number of such people and see if you can possibly make it to the exit without disturbing too many people.

Given the total number of rows and columns in the theater (nRows and nCols, respectively), and the row and column you're sitting in, return the number of people who sit strictly behind you and in your column or to the left, assuming all seats are occupied.

Example

For nCols = 16, nRows = 11, col = 5, and row = 3, the output should be
seatsInTheater(nCols, nRows, col, row) = 96.

Here is what the theater looks like:

Input/Output

• [execution time limit] 0.5 seconds (c)

• [input] integer nCols

  An integer, the number of theater's columns.

  Guaranteed constraints:
  1 ≤ nCols ≤ 1000.

• [input] integer nRows

  An integer, the number of theater's rows.

  Guaranteed constraints:
  1 ≤ nRows ≤ 1000.

• [input] integer col

  An integer, the column number of your own seat (1-based).

  Guaranteed constraints:
  1 ≤ col ≤ nCols.

• [input] integer row

  An integer, the row number of your own seat (1-based).

  Guaranteed constraints:
  1 ≤ row ≤ nRows.

• [output] integer

  • The number of people who sit strictly behind you and in your column or to the left.

[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;
}

int seatsInTheater(int nCols,int nRows,int col,int row)
{
	return (nCols-col+1)*(nRows-row);
}

Easy

n children have got m pieces of candy. They want to eat as much candy as they can, but each child must eat exactly the same amount of candy as any other child. Determine how many pieces of candy will be eaten by all the children together. Individual pieces of candy cannot be split.

Example

For n = 3 and m = 10, the output should be
candies(n, m) = 9.

Each child will eat 3 pieces. So the answer is 9.

Input/Output

• [execution time limit] 0.5 seconds (c)

• [input] integer n

  The number of children.

  Guaranteed constraints:
  1 ≤ n ≤ 10.

• [input] integer m

  The number of pieces of candy.

  Guaranteed constraints:
  2 ≤ m ≤ 100.

• [output] integer

  • The total number of pieces of candy the children will eat provided they eat as much as they can and all children eat the same amount.

[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;
}

int candies(int n,int m)
{
	return m-m%n;
}

Easy

Given an integer n, return the largest number that contains exactly n digits.

Example

For n = 2, the output should be
largestNumber(n) = 99.

Input/Output

• [execution time limit] 0.5 seconds (c)

• [input] integer n

  Guaranteed constraints:
  1 ≤ n ≤ 9.

• [output] integer

  • The largest integer of length n.

[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;
}

int largestNumber(int n)
{
	int N=0;

	for(int i=0;i<n;i++)
	{
		N*=10;
		N+=9;
	}

	return N;
}

Easy

You are given a two-digit integer n. Return the sum of its digits.

Example

For n = 29, the output should be
addTwoDigits(n) = 11.

Input/Output

• [execution time limit] 0.5 seconds (c)

• [input] integer n

  A positive two-digit integer.

  Guaranteed constraints:
  10 ≤ n ≤ 99.

• [output] integer

  • The sum of the first and second digits of the input number.

[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;
}

int addTwoDigits(int n)
{
	return n/10+n%10;
}