bobbilyking's blog

CSES Coin comb I TLE: why?
By bobbilyking, history, 5 years 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?

  • Vote: I like it
  • 0
  • Vote: I do not like it

Comments Comments (14)
| Write comment?
»
5 years ago, # |
  Vote: I like it +9 Vote: I do not like it

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

→ Reply
  • »
    »
    5 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    omg that worked ty

    → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    Had been scratching my head around what further to optimize....well there's always an optimization ty :)

    → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    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

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      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.

      → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it +7 Vote: I do not like it

      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
      → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It works, thanks.

    → Reply
»
4 years 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];
→ Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    This saved me in other problem, I was getting TLE with almost the same code I done 2 years before. just adding a const and it got accepted.

    In Problem "Graph paths I" runs 4 times faster, about 57 million of '%' operations in the worst test case.

    → Reply
  • »
    »
    15 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    For java user, declare mod globally as:

    static final int mod = 1000000007;

    and it will work fine

    → Reply
    • »
      »
      »
      4 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      can you share the code?

      cause, I have kept mod as you suggested but still getting TLE over 3 test cases!

      → Reply
»
14 months ago, # |
← Rev. 5   Vote: I like it 0 Vote: I do not like it

[deleted comment]

→ Reply