NeverMa573r's blog

A different approach to 1925B
By NeverMa573r, history, 8 months ago, In English

After looking at the hints for 1925B (https://mirror.codeforces.com/problemset/problem/1925/B), I realized the solution by knowing the properties of gcd(a,b) = gcd(a, a + b). But since this is codeforces, I'm wondering how I would have got to the solution if I hadn't known this property about gcd. For those who have solved this without using mathematical properties, how did you solve this?

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8 months ago, # |
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easy problem just look for the smallest divisor of x that is greater than or equal to n , answer is x by that number

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    8 months ago, # ^ |
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    how did you arrive to that solution? Did you just know the gcd property?

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      8 months ago, # ^ |
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      you are acting like the only one who knows euclids's lemma it is very famous upd: my sol above gives WA

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        8 months ago, # ^ |
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        ok, so you just knew the property. I'm looking for someones thought process if they didn't know it. I assume in the future, there will be problems featuring mathematical properties that not everyone knows, and I assume that there should be a way to get to the solution without knowing the properties

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          8 months ago, # ^ |
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          actually i just guessed the answer and did not use the property upd:finally AC

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            8 months ago, # ^ |
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            ah, so its just intuitive to you. But I assume you knew euclids lemma before this?

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              8 months ago, # ^ |
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              yeah i just read samples and it hit into my head i usually apply this for B's which have just very little input and are usually based on formula or any answer that is at most O(log n)

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8 months ago, # |
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Let y be the maximum gcd of all subproblems difficulties so y*a1 + y*a2 +...+y*an = x Then y(a1 +...+an) = x Then a1+ ... + an is a divisor of x, and to make y big, we should make that sum as small as possible, so we can make a1+...+ a(n-1) = n-1 and we will choose an such that the sum is a divisor of x.

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