Practice problems #

None of the following problems require a knowledge of loops, conditional statements etc.

Question 1 #

Problem Description #

Keval has forgotten the password to his I-Pad. The password is a 4-digit number where each digit takes a value from 0 to 9. The good thing is that Keval remembers that his password had exactly two unique digits, and each of these digits appeared exactly twice in the password. Keval also remembers that n digits (where, \( 0 \le n \le 10 \) ) from 0-9 were definitely not used in the password.

Find the number of different possible passwords Keval could have.

Note that the password can start with the digit 0

Input constraints #

Input format #

The only line of input contains a single integer n, the number of digits which were not used in the password

Output Format #

Output a single integer that denotes the number of possible password sequences.

Sample input and output #

Solution #

#include <stdio.h>

int main() {
    int n;
    scanf("%d", &n);
    int ans = ((10-n)*(10-n-1))/2 * 6; // (10-n)C2 * 4C2
    printf("%d\n", ans);
    return 0;
}

Question 2 #

Problem Description #

You are given two integers x and y. Output the minimum integer z, such that x | z = y. Here, “|” denotes the bitwise-OR operator

Input constraints #

Input format #

The only line of input contains two space-separated integers x and y

Output Format #

Output a single integer that denotes z as described in the problem statement

Sample input and output #

Solution #

#include <stdio.h>

int main() {
    int x, y;
    scanf("%d %d", &x, &y);

    int z = x ^ y;
    printf("%d", z);

    return 0;
}

Question 3 #

Problem Description #

Reverse a 4-digit number using the modulo operator.

Input constraints #

The input would be a valid 4-digit number (i.e. no leading zeroes would be present)

Input format #

The only line of input contains a single integer n

Output Format #

Output a single integer which denotes the reverse of the n. Note that the output may have leading zeroes.

Sample input and output #

Solution #

#include <stdio.h>

int main() {
    int n, reversed = 0;

    // Input a 4-digit number
    printf("Enter a 4-digit number: ");
    scanf("%d", &n);

    // Reverse the number using modulo operator
    reversed = (n % 10) * 1000 + ((n / 10) % 10) * 100 + ((n / 100) % 10) * 10 + (n / 1000);

    // Output the reversed number
    printf("Reversed number: %d\n", reversed);

    return 0;
}

Question 4 #

Problem Description #

Given a time as input in hh:mm:ss format, print the total number of minutes which have passed in the day.

Input format #

The only line of input would contain the time in hh:mm:ss format (24-hour clock)

Output Format #

Output the total number of minutes that have passed. Note that this may not be an integer.

Sample input and output #

Solution #

#include <stdio.h>

int main() {
    // Declare three variables that will be used to store the hours, minutes,
    // and seconds, respectively
    int h, m, s;
    
    // Take input in the specified format
    scanf("%d:%d:%d", &h, &m, &s);

    /**
     * Calculate the total minutes
     * Total minutes = hours * 60 + minutes + seconds / 60
     * However, to ensure that float division is done, we typecast the seconds
     * to float
     */
    float minutes = 60 * h + m + (float)s / 60;

    // Output the total minutes as a floating point number
    printf("%f", minutes);
}

Question 5 #

Problem Description #

Kishore is a very strange boy. He keeps solving weird problems and is now stuck on one such question. Please help him solve it.

Given an integer n, you need to find the number of ordered pairs of integers (a, b) such that, \( 1 \le a \le b \le n \) and

In this problem, gcd(a, b) denotes the greatest common divisor of the numbers a and b, and lcm(a, b) denotes their lowest common multiple.

Input constraints #

Input format #

The only line of input contains a single integer n

Output Format #

Output a single integer that denotes the number of pairs (a, b) that satisfy the given constraints

Sample input and output #

Solution #

#include <stdio.h>

int main() {
    int n;
    scanf("%d", &n);

    int ans = n + 2 * ((n / 2) + (n / 3));
    printf("%d", n);

    return 0;
}