Siriuslight's blog

By Siriuslight, history, 3 weeks ago, In English,

How to solve this problem using matrix exponentiation. The recurrence relation is :

f(n, k, 0) = 2 * f(n - 1, k, 1) + f(n - 1, k, 0)

f(n, k, 1) = f(n - 1, k, 1) + f(n - 1, k, 0)

1 < n < 1e9

1 < k < 1e3

 
 
 
 

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3 weeks ago, # |
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Matrix exponentiation is and it can be optimized to , but it's too slow and may still get TLE.

I used a O(k) solution.

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3 weeks ago, # |
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I solved it by Lagrange Interpolation

Let f(n, k) denote the answer then we have the recurrence

f(n, k) - f(n - 1, k) = f(n - 1, k - 1) + f(n - 1, k - 2)

Now $f(n, 1) = 2n $ , which is linear, which implies that f(n, 2) must be quadratic, which implies that f(n, 3) must be cubic and so on..

So for fixed k, f(n, k) will be a kth degree polynomial in n, and therefore i precomputed f(n, k) for n, k upto 2000 and then I answered each test case in O(n) using Lagrange Interpolation.

So overall time complexity O(K2 + TN)

My solution for reference

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    3 weeks ago, # ^ |
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    I have heard of this term first time. Maths is really very important for progress in CP. BTW thank you for your solution.

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3 weeks ago, # |
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It can be done by combinatorics

https://www.codechef.com/viewsolution/20517309

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3 weeks ago, # |
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Wrote a DP for N and K = 5K and fed the values of (N,1)....(N,K) into BM algo given in the blog HERE