To solve this problem, we need to compute the sum of all divrs of a given integer ( n ). Divrs of a number come in pairs, which allows us to optimize our solution by checking up to the square root of ( n ) instead of the entire range up to ( n ).

Approach

1. Edge Case Handling: If ( n ) is non-positive, return 0 since divrs of non-positive integers are either undefined or not considered here.
2. Pairwise Divrs: For each integer ( i ) from 1 to the square root of ( n ):
   • If ( i ) is a divr of ( n ), add ( i ) to the sum.
   • If ( i ) is not the same as ( n/i ) (to avoid double-counting perfect squares), add ( n/i ) to the sum.
3. Efficiency: This approach reduces the time complexity to ( O(\sqrt{n}) ), which is much faster than the naive ( O(n) ) approach.

Solution Code

def sum_of_divrs(n):
    if n <= 0:
        return 0
    total = 0
    for i in range(1, int(n ** 0.5) + 1):
        if n % i == 0:
            total += i
            if i != n // i:
                total += n // i
    return total

Explanation

• Edge Case: The function first checks if ( n ) is non-positive and returns 0 immediately.
• Loop Through Divrs: The loop runs from 1 to ( \sqrt{n} ) (inclusive). For each ( i ):
  • If ( i ) divides ( n ), it is added to the sum.
  • The corresponding pair ( n/i ) is added only if it is different from ( i ) (to avoid adding the square root twice for perfect squares).
• Result: The total sum of all divrs is returned.

This method efficiently computes the sum of divrs with optimal time complexity, making it suitable for large values of ( n ).

Examples:

• ( sum_of_divrs(6) = 1 + 2 + 3 + 6 = 12 )
• ( sum_of_divrs(4) = 1 + 2 +4 =7 )
• ( sum_of_divrs(12)=1+2+3+4+6+12=28 )

All these examples are handled correctly by the function.

Answer: The function sum_of_divrs as defined above is the solution. The final answer is the code provided.

def sum_of_divrs(n):
    if n <= 0:
        return 0
    total = 0
    for i in range(1, int(n ** 0.5) + 1):
        if n % i == 0:
            total += i
            if i != n // i:
                total += n // i
    return total

作者声明:本文包含人工智能生成内容。