Author: root

Problem Description

Find all prime numbers less than or equal to the given natural number N.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. For example, 2, 3, 5, 7, 11, 13, 17, 19 are prime numbers.

Input

A natural number N (2 ≤ N ≤ 10,000,000)

Output

Print all prime numbers less than or equal to N, one per line

Example

Input:
10

Output:
2
3
5
7

Problem Solving Strategy

One of the most well-known algorithms for finding prime numbers is the Sieve of Eratosthenes. This algorithm helps efficiently find all prime numbers within a given range.

The Sieve of Eratosthenes algorithm works as follows:

1. Create an array containing all numbers from 2 to N.
2. Since 2 is prime, remove all multiples of 2 from the array as they cannot be prime.
3. Next, select the first remaining prime number, which is 3, and remove all multiples of 3.
4. Repeat this process until the end of the array. The remaining numbers are primes.

C# Implementation

Let’s implement a program to find prime numbers in C# using the above method:

using System;

class Program
{
    static void Main(string[] args)
    {
        // Get N from the user
        Console.Write("Enter a natural number N (2 ≤ N ≤ 10,000,000): ");
        int N = int.Parse(Console.ReadLine());

        // Initialize a vector array for prime checking
        bool[] isPrime = new bool[N + 1];

        // Initialize all numbers as prime
        for (int i = 2; i <= N; i++)
        {
            isPrime[i] = true;
        }

        // Sieve of Eratosthenes algorithm
        for (int i = 2; i * i <= N; i++)
        {
            if (isPrime[i])
            {
                // Set multiples of i to false
                for (int j = i * i; j <= N; j += i)
                {
                    isPrime[j] = false;
                }
            }
        }

        // Print results
        Console.WriteLine($"The following are prime numbers less than or equal to {N}:");
        for (int i = 2; i <= N; i++)
        {
            if (isPrime[i])
            {
                Console.WriteLine(i);
            }
        }
    }
}

Code Explanation

• Getting Input: Requests the user to input N. Converts the received value to an integer.
• Initializing Prime Array: Creates a boolean array initialized to true, assuming all numbers up to N are prime.
• Implementing Sieve of Eratosthenes: Uses two nested loops to set non-prime numbers to false. The outer loop starts from 2, and the inner loop sets multiples of i to false.
• Outputting Results: Finally, finds indices in the array that are still true and outputs the corresponding prime numbers.

Complexity Analysis

The time complexity of this algorithm is O(n log log n) and its space complexity is O(n). This makes it very efficient for finding large ranges of prime numbers.

Conclusion

In this course, we introduced an algorithm for finding prime numbers and explained how to implement it using C#. The Sieve of Eratosthenes is a very efficient prime detection algorithm that often appears in actual coding tests. Understanding the concept of prime numbers and implementing the algorithm will greatly enhance your programming skills.

Problem Description

The placement of parentheses can change the calculation result of an expression. For example,
2 * 3 + 4 and (2 * 3) + 4 yield the same result, but
2 * (3 + 4) gives a different result.

The goal is to solve the problem of finding the possible minimum value based on the placement of parentheses in a given expression.
The expression consists of numbers and operators (+, -, *).

Problem Definition

Given an array of integers and operators, perform the task of appropriately placing parentheses to
produce the smallest result.
Specifically, when the expression takes the following form:

2 * 3 - 5 + 7

You need to find a way to minimize this expression using parentheses.

Problem Approach

A strategy to solve this problem is to use recursive exploration to consider all placements of parentheses.
Generate all combinations of placements and compute the result for each case to find the smallest value.

1. Simplifying the Problem

First, let’s express the following expression in an array format.
For example, 2 * 3 - 5 + 7 is transformed into the following structure:

[2, '*', 3, '-', 5, '+', 7]

2. Recursive Approach

To place parentheses controlling long expressions at each possible position,
we will use a recursive function to explore all cases.
The main steps are as follows:

• Recursively divide the expression and add parentheses.
• Calculate the result of each sub-expression.
• Update the minimum value among the calculated results.

3. Code Implementation

Below is an example code implemented in C#.

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        string expression = "2*3-5+7";
        int result = MinValue(expression);
        Console.WriteLine("Minimum Value: " + result);
    }

    static int MinValue(string expression)
    {
        var numbers = new List();
        var operators = new List();

        // Split the input string into numbers and operators.
        for (int i = 0; i < expression.Length; i++)
        {
            if (char.IsDigit(expression[i]))
            {
                int num = 0;
                while (i < expression.Length && char.IsDigit(expression[i]))
                {
                    num = num * 10 + (expression[i] - '0');
                    i++;
                }
                numbers.Add(num);
                i--; // Adjust i value
            }
            else
            {
                operators.Add(expression[i]);
            }
        }

        return CalculateMin(numbers, operators);
    }

    static int CalculateMin(List numbers, List operators)
    {
        // Base case: When only one number is left
        if (numbers.Count == 1)
            return numbers[0];

        int minValue = int.MaxValue;

        for (int i = 0; i < operators.Count; i++)
        {
            char op = operators[i];
            List leftNumbers = numbers.ToList();
            List rightNumbers = numbers.ToList();
            List leftOperators = operators.GetRange(0, i);
            List rightOperators = operators.GetRange(i + 1, operators.Count - i - 1);

            // Divide the left and right expressions based on the operator.
            int leftValue = CalculateMin(leftNumbers.GetRange(0, i + 1), leftOperators);
            int rightValue = CalculateMin(rightNumbers.GetRange(i + 1, rightNumbers.Count - i - 1), rightOperators);

            // Perform the operation.
            int result = PerformOperation(leftValue, rightValue, op);

            // Update the minimum value
            if (result < minValue)
                minValue = result;
        }

        return minValue;
    }

    static int PerformOperation(int left, int right, char op)
    {
        switch (op)
        {
            case '+':
                return left + right;
            case '-':
                return left - right;
            case '*':
                return left * right;
            default:
                throw new InvalidOperationException("Unsupported operator.");
        }
    }
}

4. Code Explanation

The functions used in the above code serve the following purposes:

• MinValue: Splits the given expression string into numbers and operators and prepares initial data for minimum value calculation.
• CalculateMin: Recursively calculates all possible sub-expressions and finds the minimum value.
• PerformOperation: Performs calculations using two numbers and an operator.

Through this structure, all combinations of parentheses placements are explored, and the minimum value among the calculated results is derived.

Conclusion

I hope this example problem has helped you understand the placement of parentheses and algorithmic approaches.
Based on this method and code, you can tackle various problems to find the minimum values of different expressions.
By always simplifying the problem and practicing recursive approaches, you can maximize your algorithmic thinking.

Additional Learning Resources

For a deeper understanding of algorithms, please refer to the following materials:

• GeeksforGeeks – Algorithm Fundamentals
• LeetCode – Algorithm Exercises
• HackerRank – Algorithms