Problem Link: http://www.spoj.com/problems/DIVSUM/
SUMMARY:
e.g. number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.
Logic
The logic for solving this question is that divisors always exist in pairs. So, for a number ‘t’, if ‘i’ is a divisor then ‘t/i’ is also a divisor.
And the smaller number of any divisor pair is always less than or equal to the square root of the number.
To View My Solution:
https://github.com/shivam04/spoj/blob/master/DIVSUM.cpp
SUMMARY:
Given a natural number n (1 <= n <= 500000), please output the summation of all its proper divisors.
Definition: A proper divisor of a natural number is the divisor that is strictly less than the number.e.g. number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.
Logic
The logic for solving this question is that divisors always exist in pairs. So, for a number ‘t’, if ‘i’ is a divisor then ‘t/i’ is also a divisor.
And the smaller number of any divisor pair is always less than or equal to the square root of the number.
To View My Solution:
https://github.com/shivam04/spoj/blob/master/DIVSUM.cpp
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