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Heisenbug's blog

By Heisenbug, 9 years ago, In English

Is there a stupid trick to calculate n-choose-k for a range of n with k always the same, which is faster than forming the full Pascal's triangle? With some fixed k, I need to calculate all n-choose-k modulo 10^9+7 for n up to 100,000 but there's no way I can calculate all the coefficients within 3 seconds.

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9 years ago, # |
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C(N, K) = N! / (K! * (N — K)!) => C(N + 1, K) = C(N, K) * (N + 1) / (N + 1 — K), that's obvious.

I think, you don't know how to perform the division, so please read about modular multiplicative inverse.

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9 years ago, # |
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You can divide by x over a prime modulus M by multiplying by xM - 2. This lets you use factorial formula for n choose k.

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9 years ago, # |
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Well, for that task you don't even need the fact that K is constant, if you precalculate all the factorials, you can answer the nCk query in logarithmic time.

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    9 years ago, # ^ |
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    Won't you answer the query in O(1) time then?

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      9 years ago, # ^ |
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      Nope, you will need to calculate modular inverse(O(logN)).

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        9 years ago, # ^ |
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        Actually, you can calculate modular inverse O(1) with O(n) pre-calculate.

        Let rv[x] is the inverse of x under M, then rv[x] = rv[M % x] * (M — M / x) % M and of course rv[1] = 1.

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          9 years ago, # ^ |
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          That's awesome!! I'm eager to know why it works. Would you explain please?

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            9 years ago, # ^ |
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            let M = kx + r, r < x, then k = M / x, r = M % x and

            both multiple rv[x] we will get

            so,