Basic Questions

Print number from 1 to n

int main() {
    int n;
    cin>>n;
    int i = 1;
    while(i<=n) {
        cout<<i<<" ";
        i = i+1;
    }
}

int main() {
    int n;
    cin>>n;
    int i = 1;
    while(i<=n) {
        cout<<i<<" ";
        i = i+1;
    }
}

int main() {
    int n;
    cin>>n;
    int i = 1;
    while(i<=n) {
        cout<<i<<" ";
        i = i+1;
    }
}

Print sum of number from 1 to n

int main() {
   int n;
    cin >> n;
    int i = 1;
    int sum = 0;
    while(i<=n) {
        sum = sum + i;
        i = i + 1;
    }
    cout<< "value of sum is "<<sum<<endl;
}

Find sum of all even number from 1 to n

int main() {
   int n;
    cin >> n;
    int i = 1;
    int sum = 0;
    while(i<=n) {
         if(i % 2 == 0) {
            sum = sum + i;  // Add 'i' to sum if it's even
        }
        i = i + 1;
    }
    cout<< "value of sum is "<<sum<<endl;
}

Convert fahrenheit to celcius

int main() {
    float fahrenheit, celsius;

    // Take input temperature in Fahrenheit from the user
    cout << "Enter temperature in Fahrenheit: ";
    cin >> fahrenheit;

    // Formula to convert Fahrenheit to Celsius
    celsius = (fahrenheit - 32) * 5 / 9;

    // Output the temperature in Celsius
    cout << "Temperature in Celsius: " << celsius << endl;

    return 0;
}

Convert celcius to fahrenheit

int main() {
    float celsius, fahrenheit;
    // Take input temperature in Celsius from the user
    cout << "Enter temperature in Celsius: ";
    cin >> celsius;
    // Formula to convert Celsius to Fahrenheit
    fahrenheit = (celsius * 9 / 5) + 32;
    // Output the temperature in Fahrenheit
    cout << "Temperature in Fahrenheit: " << fahrenheit << endl;
    return 0;
}

Find all prime number from 1 to n

int main() {
    int n;
    // Take input from the user for the value of 'n'
    cout << "Enter the value of n: ";
    cin >> n;
    int i = 2;  // Start from 2 since 1 is not a prime number
    // Loop through numbers from 2 to 'n'
    while(i <= n) {
        bool isPrime = true;  // Assume 'i' is prime
        // Check if 'i' is divisible by any number from 2 to i-1
        int j = 2;
        while(j <= i / 2) {  // Check up to i / 2 for factors
            if(i % j == 0) {
                isPrime = false;  // If divisible, it's not prime
                break;  // No need to check further
            }
            j = j + 1;
        }
        // If 'i' is still considered prime, print it
        if(isPrime) {
            cout << i << " ";
        }
        i = i + 1;  // Increment 'i' for the next iteration
    }

    return 0;
}

Check given number is prime or not

int main() {
    int n;
    bool isPrime = true;  // Assume the number is prime initially
    // Take input from the user for the value of 'n'
    cout << "Enter a number: ";
    cin >> n;
    // Edge case: numbers less than or equal to 1 are not prime
    if(n <= 1) {
        isPrime = false;
    } else {
        // Check if 'n' is divisible by any number from 2 to sqrt(n)
        for(int i = 2; i <= n / 2; i++) {
            if(n % i == 0) {
                isPrime = false;  // If divisible, it's not prime
                break;  // No need to check further
            }
        }
    }
    // Output whether the number is prime or not
    if(isPrime) {
        cout << n << " is a prime number." << endl;
    } else {
        cout << n << " is not a prime number." << endl;
    }
    return 0;
}

Print fibonacci series of numbers from 1 to n

int main() {
    int n = 10; // The number of Fibonacci numbers to generate

    int a = 0; // Initialize the first number of the Fibonacci sequence (F(0))
    int b = 1; // Initialize the second number of the Fibonacci sequence (F(1))

    // Print the first two numbers of the Fibonacci sequence
    cout << a << " " << b << " ";

    // Loop to calculate and print the next n-2 Fibonacci numbers (since the first two are printed above)
    for (int i = 1; i <= n; i++) {
        int nextNumber = a + b; // Calculate the next Fibonacci number by summing the last two numbers

        cout << nextNumber << " "; // Print the next Fibonacci number

        a = b; // Update 'a' to the previous 'b' (move one step forward in the sequence)
        b = nextNumber; // Update 'b' to the new Fibonacci number (next in the sequence)
    }

    return 0; // Indicate that the program finished successfully
}

0 1 1 2 3 5 8 13 21 34 55 89

Subtract product and sum from given number 1 to n.

// Function to calculate the difference between the product and the sum of digits of an integer
int subtractProductAndSum(int n) {
    int prod = 1; // Initialize product to 1 (multiplicative identity)
    int sum = 0;  // Initialize sum to 0 (additive identity)

    // Loop until there are no digits left in n
    while (n != 0) {
        int digit = n % 10; // Get the last digit of n (using modulus operator)
        prod = prod * digit; // Multiply the current product by the digit
        sum = sum + digit;   // Add the digit to the sum
        n = n / 10;          // Remove the last digit from n (using integer division)
    }

    int answer = prod - sum; // Calculate the difference between product and sum
    return answer;            // Return the result
}

Example If you call subtractProductAndSum(234), the steps would be as follows:

Digits: 2, 3, and 4.
Product: 2×3×4=24.
Sum: 2+3+4=9
Result: 24−9=15

Convert Binary to Decimal

int main() {
    int n; // Declare an integer variable to store the input
    cin >> n; // Read an integer from standard input

    int i = 0; // Initialize a variable to keep track of the digit's position (0 for the least significant digit)
    int ans = 0; // Initialize the answer variable to store the decimal equivalent

    // Loop until n becomes 0
    while (n != 0) {
        int digit = n % 10; // Extract the last digit of n using modulus operator
        if (digit == 1) { // Check if the extracted digit is 1
            ans = ans + pow(2, i); // If so, add 2 raised to the power of its position (i) to ans
        }
        n = n / 10; // Remove the last digit from n (using integer division)
        i++; // Increment the position counter
    }

    cout << ans << endl; // Print the final result
    return 0; // Return from the main function
}

Example: Input: 111

Process:
- First Iteration:
  - digit = 1, ans += pow(2, 0) → ans = 1
  - n = 11, i = 1
- Second Iteration:
  - digit = 1, ans += pow(2, 1) → ans = 1 + 2 = 3
  - n = 1, i = 2
- Third Iteration:
  - digit = 1, ans += pow(2, 2) → ans = 3 + 4 = 7
  - n = 0, loop ends.
Output: 7 (the decimal equivalent of binary 111).

Convert Decimal to Binary

  int main() {
    int n; // Declare an integer variable to store the input
    cin >> n; // Read an integer from standard input

    int ans = 0; // Initialize the answer variable to store the decimal equivalent
    int i = 0; // Initialize a variable to keep track of the position of the bit

    // Loop until n becomes 0
    while (n != 0) {
        int bit = n & 1; // Extract the least significant bit (LSB) using the bitwise AND operator
        ans = (bit * pow(10, i)) + ans; // Add the bit at the current position to ans
        n = n >> 1; // Right shift n to process the next bit
        i++; // Increment the position counter
    }

    cout << " Answer is " << ans << endl; // Print the final result
    return 0; // Return from the main function
}

Example: Input: 10

The binary representation of 10 is 1010. Process:

Initial state: n = 10, ans = 0, i = 0
- First Iteration:
  - bit = 10 & 1 → bit = 0
  - ans = (0 * pow(10, 0)) + 0 → ans = 0
  - n = 10 >> 1 → n = 5, i = 1
  - Current state: n = 5, ans = 0, i = 1
- Second Iteration:
  - bit = 5 & 1 → bit = 1
  - ans = (1 * pow(10, 1)) + 0 → ans = 10
  - n = 5 >> 1 → n = 2, i = 2
Current state: n = 2, ans = 10, i = 2
- Third Iteration:
  - bit = 2 & 1 → bit = 0
  - ans = (0 * pow(10, 2)) + 10 → ans = 10
  - n = 2 >> 1 → n = 1, i = 3
Current state: n = 1, ans = 10, i = 3
- Fourth Iteration:
  - bit = 1 & 1 → bit = 1
  - ans = (1 * pow(10, 3)) + 10 → ans = 1000 + 10 = 1010
  - n = 1 >> 1 → n = 0, i = 4
Loop ends since n is now 0.

Output: Answer is 1010

Reverse Integer

#include <climits> // Include limits.h for INT_MAX and INT_MIN

int reverse(int x) {
    int ans = 0; // Initialize the result variable to store the reversed number

    // Loop until x becomes 0
    while (x != 0) {
        int digit = x % 10; // Extract the last digit of x

        // Check for overflow/underflow before updating ans
        if ((ans > INT_MAX / 10) || (ans < INT_MIN / 10)) {
            return 0; // Return 0 if reversing would cause overflow/underflow
        }

        ans = (ans * 10) + digit; // Shift ans left and add the digit
        x = x / 10; // Remove the last digit from x
    }

    return ans; // Return the reversed number
}

Example:

Input: 123
Process:

Initial state: x = 123, ans = 0
- First Iteration:
  - digit = 123 % 10 → digit = 3
  - Check for overflow: ans is 0, so it’s safe.
  - ans = (0 * 10) + 3 → ans = 3
  - x = 123 / 10 → x = 12
Current state: x = 12, ans = 3
- Second Iteration:
  - digit = 12 % 10 → digit = 2
  - Check for overflow: ans is 3, so it’s safe.
  - ans = (3 * 10) + 2 → ans = 32
  - x = 12 / 10 → x = 1
Current state: x = 1, ans = 32
- Third Iteration:
  - digit = 1 % 10 → digit = 1
  - Check for overflow: ans is 32, so it’s safe.
  - ans = (32 * 10) + 1 → ans = 321
  - x = 1 / 10 → x = 0

The loop ends since x is now 0.

Output: 321

Check given number is power of 2 or not

#include <limits.h> // Include limits.h for INT_MAX

bool isPowerOfTwo(int n) {
    int ans = 1; // Initialize ans to 1, which represents 2^0

    // Loop from 0 to 30 (to cover powers of 2 from 2^0 to 2^30)
    for (int i = 0; i <= 30; i++) {
        // Check if the current power of 2 equals n
        if (ans == n) {
            return true; // Return true if n is a power of two
        }

        // Check if doubling ans will not cause overflow
        if (ans < INT_MAX / 2) {
            ans = ans * 2; // Multiply ans by 2 to get the next power of two
        }
    }
    return false; // Return false if n is not a power of two
}

Example:
Input: 8
Process:
- Iteration 0:
  - ans = 1
  - ans != 8, continue.
  - ans = 2
- Iteration 1:
  - ans = 2
  - ans != 8, continue.
  - ans = 4
- Iteration 2:
  - ans = 4
  - ans != 8, continue.
  - ans = 8
- Iteration 3:
  - ans = 8
  - ans == 8, return true.
Output: true (8 is a power of 2, specifically 2³).

Compliment of base 10 Ineteger

int bitwiseComplement(int n) {
    int m = n; // Create a copy of n to work with
    int mask = 0; // Initialize mask to zero

    // Handle the special case when n is 0
    if (n == 0)
        return 1; // The complement of 0 is 1

    // Create a mask with all bits set to 1 for the length of n
    while (m != 0) {
        mask = (mask << 1) | 1; // Shift mask left and set the least significant bit to 1
        m = m >> 1; // Right shift m to process the next bit
    }

    // Calculate the bitwise complement and apply the mask
    int ans = (~n) & mask;
    return ans; // Return the computed bitwise complement
}

Example
Input: 5 (which is 101 in binary)
Process:
1. Initial values: n = 5, m = 5, mask = 0.
2. Build the mask:
  - First Iteration:
    - mask = (0 << 1) | 1 → mask = 1 (binary 0001)
    - m = 5 >> 1 → m = 2 (binary 10)
  - Second Iteration:
    - mask = (1 << 1) | 1 → mask = 3 (binary 0011)
    - m = 2 >> 1 → m = 1 (binary 1)
  - Third Iteration:
    - mask = (3 << 1) | 1 → mask = 7 (binary 0111)
    - m = 1 >> 1 → m = 0 (binary 0)
3. Final values: mask = 7 (binary 0111).
4. Calculate the bitwise complement:
  - ~n = ~5 = -6 (binary: …11111010 for a 32-bit integer).
  - ans = (-6) & 7 = (binary -6) & (binary 0111) = (binary 010) which is 2 in decimal.

Output: 2 (the bitwise complement of 5 is 2).

Find mⁿ power

int power(int num1, int num2) {
    int ans = 1; // Initialize ans to 1, which will hold the final result
    for (int i = 1; i <= num2; i++) { // Loop from 1 to num2
        ans = ans * num1; // Multiply ans by num1 in each iteration
    }
    return ans; // Return the final result
}

Example 1

Input:

num1 = 2
num2 = 3

Process:

Initialize ans = 1.
Loop from i = 1 to 3:
- Iteration 1 (i = 1):
  - ans = 1 * 2 = 2
- Iteration 2 (i = 2):
  - ans = 2 * 2 = 4
- Iteration 3 (i = 3):
  - ans = 4 * 2 = 8
The final value of ans is 8.

Output:

8 (since ( 2^3 = 8 )).

Calculate Combination (nCr) Using Factorial

int factorial(int n) {
    int fact = 1; // Initialize the factorial result to 1
    for (int i = 1; i <= n; i++) { // Loop from 1 to n
        fact = fact * i; // Multiply fact by i to calculate the factorial
    }
    return fact; // Return the computed factorial
}

int nCr(int n, int r) {
    int num = factorial(n); // Calculate n!
    int denom = factorial(r) * factorial(n - r); // Calculate r! * (n-r)!
    return num / denom; // Return n! / (r! * (n-r)!)
}

int main() {
    int n, r; // Declare variables for n and r
    cin >> n >> r; // Input values for n and r
    cout << "Answer is " << nCr(n, r) << endl; // Output the result of nCr
    return 0; // End of the program
}

Example 1

Input:

n = 5
r = 2

Process:

Calculate Factorials:
- Calculate ( n! ):
  - factorial(5):
    - fact = 1
    - Iteration 1: fact = 1 * 1 = 1
    - Iteration 2: fact = 1 * 2 = 2
    - Iteration 3: fact = 2 * 3 = 6
    - Iteration 4: fact = 6 * 4 = 24
    - Iteration 5: fact = 24 * 5 = 120
  - Result: 5! = 120
- Calculate ( r! ):
  - factorial(2):
    - fact = 1
    - Iteration 1: fact = 1 * 1 = 1
    - Iteration 2: fact = 1 * 2 = 2
  - Result: 2! = 2
- Calculate ( (n - r)! ):
  - factorial(3) (since ( n - r = 5 - 2 = 3 )):
    - fact = 1
    - Iteration 1: fact = 1 * 1 = 1
    - Iteration 2: fact = 1 * 2 = 2
    - Iteration 3: fact = 2 * 3 = 6
  - Result: 3! = 6

Calculate ( nCr ):

Using the formula:

<Math>
  C(5, 2) = \frac{5!}{2!(5 - 2)!} = \frac{120}{2 \times 6} = \frac{120}{12} = 10
</Math>

Output:

Answer is 10

Check given number is even or odd

bool isEven(int a) {
    //odd
    if (a & 1) {
        return 0; // Return false (0) if the number is odd
    }
    else { // Even
        return 1; // Return true (1) if the number is even
    }
}

Example:

Input:

a = 4

Execution:

Perform the bitwise operation:
- ( a & 1 ) → ( 4 & 1 ) → 0 (since 4 in binary is 100, and 1 is 001)
Evaluate the condition:
- The condition if (a & 1) evaluates to false, so the function returns 1.

Output:

1 (indicating that 4 is even)

Print count numbers for n

void printCounting(int num) {
//    cout << n << endl;
    //Function Body
    for(int i=1; i<=num; i++) {
        cout<< i << " ";
    }
    cout<<endl;
}

Make a basic calculator

 int a, b;
    cout <<" Enter the value of a "<<endl;
    cin >> a;
    cout<<" Enter the value of b " <<endl;
    cin >> b;
    char op;
    cout<<" Enter the Operation you want to perform" <<endl;
    cin >> op;
    switch( op ) {
        case '+':  cout << (a+b) <<endl;
                  break;
        case '-': cout<< (a-b) <<endl;
                  break;
        case '*': cout<< (a*b) <<endl;
                  break;
        case '/': cout<< (a/b) <<endl;
                  break;
        case '%': cout<< (a%b) <<endl;
                  break;
        default: cout << "Please enter a valid Operation " << endl;

    }

Basic Questions

Print number from 1 to n

Print sum of number from 1 to n

Find sum of all even number from 1 to n

Convert fahrenheit to celcius

Convert celcius to fahrenheit

Find all prime number from 1 to n

Check given number is prime or not

Print fibonacci series of numbers from 1 to n

Subtract product and sum from given number 1 to n.

Convert Binary to Decimal

Convert Decimal to Binary

Reverse Integer

Check given number is power of 2 or not

Compliment of base 10 Ineteger

Find mn power

Example 1

Calculate Combination (nCr) Using Factorial

Example 1

Check given number is even or odd

Example:

Print count numbers for n

Make a basic calculator

Find mⁿ power