bobbilyking's blog

By bobbilyking, history, 16 months ago, In English

https://pastebin.com/zV3jsqy7

i did look at the CF editorial, and yeah my implementation (i don't think) is wrong. and someone else had the same TLE problem as me but he never said how he resolved it (if he ever did). Is this just a java thing? Editorial says that time complexity is n * target, which is 10^8 operations, so maybe I can make very very slight optimizations somewhere to get it under 1ms?

 
 
 
 
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16 months ago, # |
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I also got TLE with recursive dp in c++ and had to do iterative. Timelimit is strict but 10^8 is indeed the intended complexity. Try swapping out the modding line with this as mod is a heavy operation.

if (whatever >= MOD) whatever -= MOD

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    16 months ago, # ^ |
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    omg that worked ty

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    13 months ago, # ^ |
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    Had been scratching my head around what further to optimize....well there's always an optimization ty :)

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    4 months ago, # ^ |
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    woahhh, it really worked!! But I found another code online and it makes the use of MOD but it didn't get any TLE, how is that possible. LINK: https://usaco.guide/problems/cses-1635-coin-combinations-i-unordered/solution

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      4 months ago, # ^ |
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      I don't remember if that optimization was required in C++ codes for this problem or not. If it was, I have no idea tbh.

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      4 months ago, # ^ |
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      Btw one thing to note about this problem. You don't even need $$$n \cdot x$$$ mod calls.

      Here is the solution with n * x mod calls
      Here is the solution with x mod calls
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4 months ago, # |
Rev. 2   Vote: I like it +3 Vote: I do not like it

Not sure in java but in C++ a%b works slightly faster if b is a constant. So in the following code int mod = 1e9 + 7 will give TLE but int const mod = 1e9 + 7 will not.

dp[0] = 1;
    for (int i = 1; i <= m; i++)
    {   for (int j = 1; j<= n; j++)
        {   int x = i - a[j];
            if (x >= 0)
                dp[i] = (dp[i] + dp[x])%mod;
        }
    }
    cout << dp[m];