red_coder's blog

fastest factorial....
By red_coder, 13 years ago, In English

suppose for any 'n' taken as input we want to output (2n)!/n!(n+1)! modulo 1000000007 then which is the fastest way to accomplish this task......

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13 years ago, # |
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You can try to find a recurrence relation for your formula, then use matrix multiplication method to calculate Nth element of the sequence described by that recurrence relation.

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    13 years ago, # ^ |
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    Can you calculate this faster than ?

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      13 years ago, # ^ |
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      And how to calculate it for n / (log^2(n)) ?

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      13 years ago, # ^ |
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      I'm not sure whether it is possible or not, but for example let's take the vector V = (n!, 0) and matrix M.

      M =

      V × M(n + 1) = V0 = (n!, (n + 1)!)

      Let's take another matrix P.

      P =

      V0 × P = ((n + 1)!, 0)

      So it means that (n!, 0) × M(n + 1) × P = ((n + 1)!, 0)

      I mean to use something similar to multiply by matrix in logN time.

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        13 years ago, # ^ |
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        I think it's not possible, because the formula is non linear.

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          13 years ago, # ^ |
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          Actually , which is not linear, but it can be represented in the following linear recurrence relation:

          C(n, k) = C(n - 1, k) + C(n - 1, k - 1)

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            13 years ago, # ^ |
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            I mean if we talk about Fibonacci numbers we have vector (f[n], f[n — 1]) and the next vector can be obtained multiplying it with constant matrix. In this case we can have (n!, n) but next vector is (n!*(n+1), n+1). The second element is OK, but to produce first one we need to multiply v[0] and (v[1] + 1). That is what i call non-linear formula.

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              13 years ago, # ^ |
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              But there is a way to calculate C(n, i) in O(N) time.

              where r — row, c --column

              source: wikipedia

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                13 years ago, # ^ |
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                n! can be calculated in O(n). congratulations.

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                  13 years ago, # ^ |
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                  Not n! I mean v_c = \choose{n}{c}

                  So, we will get C(n, 0), C(n, 1), C(n, 2), ..., C(n, n) in O(N) time. Without using pascal's triangle or smth else.

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                  13 years ago, # ^ |
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                  c[n][k] = n!/((n-k)!k!).

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13 years ago, # |
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(2n)!/n!(n+1)! = 2^n / (n + 1). 2^n requires O(log n) time and 1 / (n + 1) % M is simply (n + 1)^(M — 2) % M (where M = 1000000007) because M is prime. So the total complexity is log M + log n which is fast enough I think

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    13 years ago, # ^ |
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      13 years ago, # ^ |
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      Ops.. I didn't mean that but you're right, my formula is wrong

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13 years ago, # |
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Btw, (2n)!/n!(n+1)! is the n-th Catalan number. So the problem of it's calcilation must be well known. Am I wrong?

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13 years ago, # |
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(2n)!/n!(n+1)!= c(2n,n)/(n+1)= c(2n,n)*(n+1)^(1000000005) Is it enough for you?:)

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    13 years ago, # ^ |
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    There isn't way to calculate C(2n,n) fast, is it?

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13 years ago, # |
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I think there is the way to precalculate some numbers, and then answer the queries fast enough. There is only one test case? or many queries to find n-th Catalan number? we need to find all prime numbers between 1 and mod, and some additional information.

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