{"post":"

Understanding Prime Numbers

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The number 1 is neither prime nor composite.

Examples of prime numbers:

2, 3, 5, 7, 11, 13

Solution Approach

To find all prime numbers between 1 and 100:

1. Start checking from 2 (the first prime number)

2. For each number i from 2 to 100:

- Check for divisors from 2 up to √i

- If no divisors found, the number is prime

Key Insight:

We only need to check divisors up to √i because:

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If i has a factor greater than √i, it must have a corresponding factor smaller than √i
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This optimization significantly reduces the number of checks needed
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Algorithm

1. Initialize outer loop from i = 2 to i = 100

2. For each i:

- Set counter ctr = 0

- Inner loop from j = 2 to j ≤ √i:

- If i % j == 0, set ctr = 1 and break

3. After inner loop:

- If ctr == 0, print i (prime number)

- Else, continue to next number

Implementation

\n#include <iostream>\n\n#include <cmath>\n\nusing namespace std;\n\nint main() {\n\n    cout << \"Prime numbers between 1 and 100:\\n\";\n\n    for(int i = 2; i <= 100; ++i) {\n\n        bool isPrime = true;\n\n        for(int j = 2; j <= sqrt(i); ++j) {\n\n            if(i % j == 0) {\n\n                isPrime = false;\n\n                break;\n\n            }\n\n        }\n\n        if(isPrime) {\n\n            cout << i << \" \";\n\n        }\n\n    }\n\n    return 0;\n\n}\n\n

Output

\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97\n\n

Improvements Made

1. Variable Naming: Changed ctr to more descriptive isPrime boolean

2. Optimization: Used sqrt(i) as the upper bound for divisor checking

3. Readability: Improved code formatting and comments

4. Output Format: Cleaned up the output presentation

5. Explanation: Clarified the mathematical reasoning behind checking up to √i

Time Complexity

The algorithm has a time complexity of O(n√n), which is efficient for the given range (1 to 100). For larger ranges, more sophisticated algorithms like the Sieve of Eratosthenes would be more efficient.

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