C++ Control Flow - Programs, Exercise & Assignements

Find the largest of three numbers

Description: Write largest(int a, int b, int c). Test with (12, 45, 30) and (7, 7, 3).

Hint:

Compare with >= and &&, or call std::max with a list.

Solution

#include <algorithm>
#include <iostream>

int largest(int a, int b, int c) {
    if (a >= b && a >= c) return a;
    if (b >= c) return b;
    return c;
}

int main() {
    std::cout << largest(12, 45, 30) << '\n';
    std::cout << largest(7, 7, 3) << '\n';
    std::cout << std::max({12, 45, 30}) << '\n';      // the library way
}

Output

45
7
45

How it works

If a is not the biggest, the answer is b or c, so one more check is enough.

Read C++ if else.

Check if a year is a leap year

Description: Write is_leap(int year). Test 1900, 2000, 2024 and 2026.

Hint:

Use % and the three rules.

Solution

#include <iostream>

bool is_leap(int year) {
    return (year % 4 == 0 && year % 100 != 0) || year % 400 == 0;
}

int main() {
    std::cout << std::boolalpha;
    for (int y : {1900, 2000, 2024, 2026}) {
        std::cout << y << " -> " << is_leap(y) << '\n';
    }
}

Output

1900 -> false
2000 -> true
2024 -> true
2026 -> false

How it works

1900 is divisible by 100 but not by 400, so it is not a leap year. 2000 is divisible by 400, so it is.

Read C++ Operators.

FizzBuzz in C++

Description: Print 1 to 15. For multiples of 3 print Fizz, for multiples of 5 Buzz, for both FizzBuzz.

Hint:

Check the multiples of 15 first.

Solution

#include <iostream>

int main() {
    for (int i = 1; i <= 15; i++) {
        if (i % 15 == 0)      std::cout << "FizzBuzz\n";
        else if (i % 3 == 0)  std::cout << "Fizz\n";
        else if (i % 5 == 0)  std::cout << "Buzz\n";
        else                  std::cout << i << '\n';
    }
}

Output

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz

How it works

The most special case (15) must be checked first, or it would print only Fizz.

Read C++ for Loop.

Sum of the digits of a number

Description: Write digit_sum(int n). For 9475 the answer is 25.

Hint:

n % 10 is the last digit, n / 10 removes it.

Solution

#include <iostream>

int digit_sum(int n) {
    int sum = 0;
    while (n > 0) {
        sum += n % 10;
        n /= 10;
    }
    return sum;
}

int main() {
    std::cout << digit_sum(9475) << ' ' << digit_sum(100) << ' ' << digit_sum(0) << '\n';
}

Output

25 1 0

How it works

The loop adds the last digit and cuts it off, until the number is 0.

Read C++ while Loop.

Print a multiplication table

Description: Print 7 x 3 = 21 for 1 to 10.

Hint:

A for loop and a format with a width, like {:2}.

Solution

#include <print>

int main() {
    int n = 7;
    for (int i = 1; i <= 10; i++) {
        std::println("{} x {:2} = {:2}", n, i, n * i);
    }
}

Output

7 x 1 = 7
7 x 2 = 14
7 x 3 = 21
7 x 4 = 28
7 x 5 = 35
7 x 6 = 42
7 x 7 = 49
7 x 8 = 56
7 x 9 = 63
7 x 10 = 70

How it works

{:2} uses at least 2 places, so the numbers line up.

Read C++ for Loop.

Check prime numbers and list the primes up to 30

Description: Write is_prime(int n) and print the primes from 1 to 30.

Hint:

Test divisors while i * i <= n.

Solution

#include <iostream>

bool is_prime(int n) {
    if (n < 2) return false;
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) return false;
    }
    return true;
}

int main() {
    for (int n = 1; n <= 30; n++) {
        if (is_prime(n)) std::cout << n << ' ';
    }
    std::cout << '\n';
}

Output

2 3 5 7 11 13 17 19 23 29

How it works

If n has a divisor bigger than its square root, it also has a smaller one, so we can stop early.

Read C++ Functions.