Mathematical Caclulation Problem

Given the formula for distance between a point and a line, we can simpliyf the calculation to |x-y| * 50:

#include <iostream>
#include <cmath>

int main() {
    int x, y;
    std::cin >> x >> y;
    std::cout << abs(y - x) * 50 << '\n';
    return 0;
}

String Output Problem

Simple string output challenge:

#include <iostream>

int main() {
    std::cout << "To iterate is human, to recurse divine.\n";
    return 0;
}

Contact List Management

Implementing a contact list with search functionality:

#include <iostream>
#include <vector>
#include <string>

struct Contact {
    std::string name;
    std::string phone;
    std::string email;
    std::string address;
    std::string notes;
};

int main() {
    int n;
    std::cin >> n;
    std::vector<Contact> contacts(n);
    
    for(auto& c : contacts) {
        std::cin >> c.name >> c.phone >> c.email >> c.address >> c.notes;
    }
    
    int queries;
    std::cin >> queries;
    while(queries--) {
        int index;
        std::cin >> index;
        if(index >= 0 && index < n) {
            auto& c = contacts[index];
            std::cout << c.name << ' ' << c.address << ' ' << c.notes << ' ' 
                      << c.email << ' ' << c.phone << '\n';
        } else {
            std::cout << "Not Found\n";
        }
    }
    return 0;
}

Basic Arithmetic Operations

Performing fundamental arithemtic operations with formatted output:

#include <iostream>
#include <iomanip>

int main() {
    int a, b;
    std::cin >> a >> b;
    std::cout << a << " + " << b << " = " << a + b << '\n';
    std::cout << a << " - " << b << " = " << a - b << '\n';
    std::cout << a << " * " << b << " = " << a * b << '\n';
    
    if(b == 0) {
        std::cout << a << " / " << b << " = Error\n";
    } else if(a % b == 0) {
        std::cout << a << " / " << b << " = " << a / b << '\n';
    } else {
        std::cout << a << " / " << b << " = " << std::fixed 
                  << std::setprecision(2) << (double)a / b << '\n';
    }
    return 0;
}

Unique Digit Year Calculation

Finding years with a specified number of unique digits:

#include <iostream>
#include <set>
#include <iomanip>

int countUniqueDigits(int year) {
    std::set<int> digits;
    if(year < 1000) digits.insert(0);
    while(year > 0) {
        digits.insert(year % 10);
        year /= 10;
    }
    return digits.size();
}

int main() {
    int start_year, target_count;
    std::cin >> start_year >> target_count;
    
    for(int year = start_year; ; year++) {
        if(countUniqueDigits(year) == target_count) {
            std::cout << year - start_year << ' ' 
                      << std::setw(4) << std::setfill('0') << year << '\n';
            break;
        }
    }
    return 0;
}