omkar21_pro's blog

Got WA because of ans*= j ,but ac with ans = ans*j
By omkar21_pro, history, 19 months ago, In English

In recent codeforces div 2 contest I got wa because of this in problem C — counting orders

I don's understand what is difference between two ,Aren't they both same ?

Here is my submission for which got wa

https://codeforces.net/contest/1828/submission/205934626

Here is same submission for which got ac by changing ans = ans* something

https://codeforces.net/contest/1828/submission/205934493

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19 months ago, # |
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Both of them are already accepted??? you have provided the same link twice

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19 months ago, # |
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Learn modular arithmetic properly. While you did ans *= something % MOD; in your WA submission it gets overflow because multiplication operation doing first. In your AC submission you did ans = ans * something % MOD; Here "something" is reducing first then it multiplying. Hope you get it.

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    19 months ago, # ^ |
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    oh Now I got it , Thank you very much

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      19 months ago, # ^ |
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      there's my submission 205933688 i think i have done the same thing but i don't know how it is giving integer overflow to your's!!

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        19 months ago, # ^ |
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        i think you should multiply first and then take the mod in different lines maybe that would work

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        19 months ago, # ^ |
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        I guess when I do ans*= something %M , Modulo operation is not performed only multiplication performed , Bad that I could not solve problem c because of this small error

        Congratulation on being specialist

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          19 months ago, # ^ |
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          int ans = 85 ; int something = 445 ; ans *= something % 100 ; cout<< ans<< el;

          OUTPUT : 3825

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            19 months ago, # ^ |
            ← Rev. 3   Vote: I like it -9 Vote: I do not like it

            Sorry, the previous statement was wrong. This part is correct though.

            If you really want to be careful then you should just do : ans = (ans%mod) * (something%mod); ans %= mod;

            This is because ans and something should both be less than mod when they are being multiplied if this is not the case than : ans > mod, something > mod, ans * something > mod*mod. This means it is > 1e18, which can overflow.

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          19 months ago, # ^ |
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          Thanks

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19 months ago, # |
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ㅤ

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19 months ago, # |
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The other comments are wrong. * and % have the same precedence. The left-most operator takes priority. Check this link.

Your line works (ans = ans * (max ( 0LL , k - sub)) % MOD;) because the last operation done is %, so at the end of the operation ans < 1e9 + 7.

This (ans *= (max ( 0LL , k - sub)) % MOD;) doesn't work because *= has a lower priority than %, so at the end of the operation we can't guarantee ans < 1e9 + 7.

This (ans = ans * (max ( 0LL , k - sub)) % MOD;) would have worked even if ans was declared as an integer, because the * operation is performed between a ll (result of std::max) and an int (ans) that will be casted to ll. The result of % will also be ll, but since it is lower than 1e9 + 7, it will be casted correctly back to int when attributing it back to ans.

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19 months ago, # |
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when you have to modulo the value, you can write it as ans*=j%mod, rather you will have to write it as ans=(ans*j)%mod;

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19 months ago, # |
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You should use (ans *= k) %= MOD;.

That's the correct code for multiplication and moduling.

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