$course$/top/OnlineCompiler.io C++ (English) Free CodeRunner questions from OnlineCompiler.io (https://onlinecompiler.io/coderunner-question-packs), licensed under CC BY 4.0. They use CodeRunner's built-in question types, so they run on any Jobe server.

]]> Write a program that reads the width and the height of a rectangle and prints its area and its perimeter on one line, separated by one space, each with exactly two digits after the decimal point.

The input is two positive real numbers, separated by a space or a new line and written with a decimal point (for example 2.5). Hint: fixed and setprecision from <iomanip>.

]]> 1.0000000 0.0000000 0 cpp_program 0 1 10, 20, ... 0 0 0 15 100 #include using namespace std; int main() { double width, height; cin >> width >> height; cout << fixed << setprecision(2); cout << width * height << " " << 2 * (width + height) << endl; } ]]> 1 0 onlinecompiler.io cpp io Write a function int days_in_month(int month, int year) that returns the number of days in the given month of the given year, where month goes from 1 (January) to 12 (December). February has 29 days in a leap year and 28 otherwise. A year is a leap year if it is divisible by 4, except years divisible by 100 that are not divisible by 400: 2000 was a leap year, 1900 was not.

1 ≤ month ≤ 12 and 1 ≤ year ≤ 9999. A switch on the month fits this well.

]]> 1.0000000 0.0000000 0 cpp_function 0 1 10, 20, ... 0 0 0 15 100 1 0 onlinecompiler.io cpp conditionals Write a function int triangle_type(int a, int b, int c) that receives three side lengths and returns:

0 if they can't form a triangle: a side is zero or negative, or a side is greater than or equal to the sum of the other two;
1 if the triangle is equilateral (all three sides equal);
2 if it is isosceles (exactly two sides equal);
3 if it is scalene (all sides different).

Each argument is between −10⁹ and 10⁹.

]]> 1.0000000 0.0000000 0 cpp_function 0 1 10, 20, ... 0 0 0 15 100 = b + c || b >= a + c || c >= a + b) { return 0; } if (a == b && b == c) { return 1; } if (a == b || b == c || a == c) { return 2; } return 3; } ]]> 1 0 onlinecompiler.io cpp conditionals Start from a positive integer n and repeat this step until n is 1: if n is even, divide it by 2; if it is odd, replace it with 3n + 1. Write a function int collatz_steps(long long n) that returns the number of steps needed. For example, 6 needs 8 steps (6, 3, 10, 5, 16, 8, 4, 2, 1), and 1 needs 0.

1 ≤ n ≤ 10⁶. Along the way the value can grow beyond the range of int, which is why the parameter is a long long.

]]> 1.0000000 0.0000000 0 cpp_function 0 1 10, 20, ... 0 0 0 15 100 1 0 onlinecompiler.io cpp loops Write a program that reads integers until it reads a 0, then prints the largest and the smallest of the numbers that came before the 0, separated by one space. The 0 only marks the end of the data: it is not one of the numbers, and anything after it is ignored.

At least one nonzero number comes before the 0. The numbers can be negative and are between −10⁹ and 10⁹.

]]> 1.0000000 0.0000000 0 cpp_program 0 1 10, 20, ... 0 0 0 15 100 using namespace std; int main() { int x; cin >> x; int largest = x; int smallest = x; while (cin >> x && x != 0) { if (x > largest) { largest = x; } if (x < smallest) { smallest = x; } } cout << largest << " " << smallest << endl; } ]]> 1 0 onlinecompiler.io cpp loops Write a function void sort_three(int& a, int& b, int& c) that rearranges the values of its three arguments so that a ≤ b ≤ c. The parameters are references, so the function changes the caller's variables. Use comparisons and swaps (swap from the standard library is fine) rather than putting the values in a vector.

The values can be negative and can repeat.

]]> 1.0000000 0.0000000 0 cpp_function 0 1 10, 20, ... 0 0 0 15 100 b) { swap(a, b); } if (b > c) { swap(b, c); } if (a > b) { swap(a, b); } } ]]> 1 0 onlinecompiler.io cpp functions In C++ several functions can share a name as long as their parameter lists differ; this is called overloading. Write two functions named area:

double area(double radius) returns the area of a circle, πr²;
double area(double width, double height) returns the area of a rectangle.

Use π with full double precision (for example acos(-1.0), or 3.14159265358979), not 3.14. The arguments are 0 or positive. The tests print the results with two digits after the decimal point.

]]> 1.0000000 0.0000000 0 cpp_function 0 1 10, 20, ... 0 0 0 15 100 1 0 onlinecompiler.io cpp functions In the Tower of Hanoi puzzle, n disks of different sizes are stacked on one peg, largest at the bottom. The goal is to move the whole stack to another peg, moving one disk at a time and never putting a disk on top of a smaller one; the third peg serves as a helper.

Write a recursive function void hanoi(int n, int from, int to, int via) that prints the moves of the shortest solution for moving n disks from peg from to peg to using peg via. Print one move per line as two numbers separated by a space: the peg the disk leaves and the peg it goes to. For example, hanoi(2, 1, 3, 2) prints:

1 2
1 3
2 3

The pegs are numbered 1, 2 and 3, and 0 ≤ n ≤ 10. For n = 0 nothing is printed.

]]> 1.0000000 0.0000000 0 cpp_function 0 1 10, 20, ... 0 0 0 15 100 1 0 onlinecompiler.io cpp recursion Write a program that reads one line and prints its words in reverse order, separated by single spaces.

]]> 1.0000000 0.0000000 0 cpp_program 0 1 10, 20, ... 0 0 0 15 100 #include #include #include using namespace std; int main() { string line, word; getline(cin, line); istringstream words(line); vector all; while (words >> word) { all.push_back(word); } for (size_t i = all.size(); i > 0; i--) { cout << all[i - 1] << (i > 1 ? " " : ""); } cout << endl; } ]]> 1 0 onlinecompiler.io cpp strings Write a function string compress(const string& s) that replaces each run of equal consecutive characters in s with the character followed by the length of the run. For example, "aaabccdddd" becomes "a3b1c2d4": every run gets its number, even a run of one. An empty string gives an empty string.

s contains only letters, and case matters (a and A are different letters). It has at most 1000 characters, so a run can be longer than 9.

]]> 1.0000000 0.0000000 0 cpp_function 0 1 10, 20, ... 0 0 0 15 100 1 0 onlinecompiler.io cpp strings Write a function double average(const vector<int>& v) that returns the average of the numbers in v, or 0 if v is empty.

The tests print the result with cout <<, so 2.5 shows as 2.5 and 3.0 as 3.

]]> 1.0000000 0.0000000 0 cpp_function 0 1 10, 20, ... 0 0 0 15 100 & v) { if (v.empty()) { return 0; } long long sum = 0; for (int x : v) { sum += x; } return static_cast(sum) / v.size(); } ]]> 1 0 onlinecompiler.io cpp lists Write a function vector<int> remove_duplicates(const vector<int>& v) that returns a new vector with the values of v, keeping only the first occurrence of each value, in the order in which the values first appear. For example, {3, 1, 3, 2, 1} gives {3, 1, 2}.

v has at most 1000 elements and can be empty. Don't sort: the result must keep the original order.

]]> 1.0000000 0.0000000 0 cpp_function 0 1 10, 20, ... 0 0 0 15 100 remove_duplicates(const vector& v) { vector result; for (int x : v) { if (find(result.begin(), result.end(), x) == result.end()) { result.push_back(x); } } return result; } ]]> 1 0 onlinecompiler.io cpp lists Write a function map<string, int> word_count(const string& text) that returns how many times each word appears in text. Words are separated by one or more spaces, and case matters: Ana and ana are different words. An empty text gives an empty map.

The test program doesn't include <map>, so start your answer with #include <map>, plus any other header you use (such as <sstream>). The tests print every entry of the map in the map's own order, which is increasing order of the keys comparing character codes (so uppercase comes before lowercase).

]]> 1.0000000 0.0000000 0 cpp_function 0 1 10, 20, ... 0 0 0 20 100 #include map word_count(const string& text) { map counts; istringstream words(text); string word; while (words >> word) { counts[word]++; } return counts; } ]]> 1 0 m = word_count("a b a c b a"); for (const auto& [word, n] : m) cout << word << " " << n << endl]]> m = word_count(" sol mar sol "); for (const auto& [word, n] : m) cout << word << " " << n << endl]]> m = word_count("ana Ana ana"); for (const auto& [word, n] : m) cout << word << " " << n << endl]]> m = word_count("1 22 1 333 22 1"); for (const auto& [word, n] : m) cout << word << " " << n << endl]]> onlinecompiler.io cpp maps Write a class Account for a simple bank account that holds a whole number of money units. It has:

a constructor Account(int initial = 0) that opens the account with balance initial (0 or more);
void deposit(int amount), which adds amount to the balance and does nothing if amount is 0 or negative;
bool withdraw(int amount), which subtracts amount and returns true only if amount is positive and not more than the balance; otherwise it changes nothing and returns false;
int balance() const, which returns the current balance.

Keep the balance in a private member, so the only way to change it is through deposit and withdraw. Balances stay below 10⁹. The tests print bool results as 1 or 0.

]]> 1.0000000 0.0000000 0 cpp_function 0 1 10, 20, ... 0 0 0 30 100 0) { balance_ += amount; } } bool withdraw(int amount) { if (amount <= 0 || amount > balance_) { return false; } balance_ -= amount; return true; } int balance() const { return balance_; } private: int balance_; }; ]]> 1 0 onlinecompiler.io cpp classes Write a class Clock that keeps a time of day on a 24-hour clock. It has:

a constructor Clock(int hours, int minutes), with 0 ≤ hours ≤ 23 and 0 ≤ minutes ≤ 59;
void add_minutes(int minutes), which moves the clock forward by minutes, or back if minutes is negative, wrapping past midnight as many times as needed (|minutes| ≤ 10⁶);
string str() const, which returns the time as HH:MM, with two digits for each part (for example 09:05).

Hint: storing the time as the number of minutes since midnight makes add_minutes short.

]]> 1.0000000 0.0000000 0 cpp_function 0 1 10, 20, ... 0 0 0 30 100 1 0 onlinecompiler.io cpp classes