So basically we can pair up the divisors for any given $n$. So, l will have the smaller one and h will store the larger one ( such that $l \times h =n$ ) and we are incrementing l by one each time and correspondingly decreasing h. So h will decrease much faster and we would be skipping lot of numbers which anyway we need not check.
PS: I forgot to include the line to check whether $l= h$ that is if n is a perfect square.
@Sudeep Salgia
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We can bring down $\frac{n}{2}$ to $\lfloor\sqrt{n}\rfloor$. I still can't understand how will it help while dealing with higher numbers. Let me compile.
Big numbers cannot be stored in long. Therefore we need something which is even bigger, like float or double. The problem with using "%" function with float/double is that it will not work. "%" is meant to work only with integers(int or long or short), It will not work with float and double as they are decimal values. Does 100%2.3 make sense?
Hence we use a workaround to find the true factors of n by using floor.
@Raghav Vaidyanathan
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Ouch. I didn't notice that you used double. But then why not use long long? long long has a higher range than a double for the same memory, since a double can only hold 15 significant digits whereas long long can hold upto 18.
Note I don't use C++ so I have no idea if long long is used commonly.
Try long double it has the max and min values $10^{4932}\quad and \quad 10^{-4932}$ respectively and has a memory of 10 bytes. Its the biggest as far as I know.
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Top NewestYou can firstly improve by running the loop till $i \leq \frac{n}{2}$. Or you can try this:
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Yes, I noticed $\frac{n}{2}$ thing. Can you explain algorithm you used? Thanks.
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So basically we can pair up the divisors for any given $n$. So, l will have the smaller one and h will store the larger one ( such that $l \times h =n$ ) and we are incrementing l by one each time and correspondingly decreasing h. So h will decrease much faster and we would be skipping lot of numbers which anyway we need not check.
PS: I forgot to include the line to check whether $l= h$ that is if n is a perfect square.
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$\frac{n}{2}$ to $\lfloor\sqrt{n}\rfloor$. I still can't understand how will it help while dealing with higher numbers. Let me compile.
We can bring downLog in to reply
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Try this out
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Why are you taking
to check if $i$ is a factor of $n$ instead of
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Big numbers cannot be stored in long. Therefore we need something which is even bigger, like float or double. The problem with using "%" function with float/double is that it will not work. "%" is meant to work only with integers(int or long or short), It will not work with float and double as they are decimal values. Does 100%2.3 make sense?
Hence we use a workaround to find the true factors of n by using floor.
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Note I don't use C++ so I have no idea if long long is used commonly.
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Try Java Codes,They are always better than C++
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For that, I'll have to learn java properly.
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Maybe you could try using long variable or long long . But that won't be really necessary if you are writing this for boards
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I program for fun. -_- I hate boards
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Oh ! So have you studied any algorithm books like CLRS ? Or are you active on websites like topcoder , spoj ?
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Try long double it has the max and min values $10^{4932}\quad and \quad 10^{-4932}$ respectively and has a memory of 10 bytes. Its the biggest as far as I know.
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Modulus doesn't work with floating data types.
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How many times should i try to help before someone notices Pranjal Jain . I have included a few ideas in the following from Sudeep Salgia as well:
Here are the factors of $10000000099$
And here are the factors of $7296872389761$
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@Pranjal Jain @Sudeep Salgia
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Thanks. Sorry for not noticing. I am quite inactive now a days on B'ant. (Turned off all email notifications as well)
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