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    <p>A word of caution: In the case c=2, len1=1, len2=2, the answer will be the N'th Fibonacci number, and the Fibonacci numbers grow (approximately) exponentially with a growth factor of the golden ratio, phi ~ 1.61803399. For the huge value N=10^15, the answer will be about phi^(10^15), an enormous number. The answer will have storage requirements on the order of (ln(phi^(10^15))/ln(2)) / (8 * 2^40) ~ 79 terabytes. Since you can't even <em>access</em> 79 terabytes in 20 seconds, it's unlikely you can meet the speed requirements in this special case.</p> <p>Your best hope occurs when C is not too large, and leni is large for all i. In such cases, the answer will still grow exponentially with N, but the growth factor may be much smaller.</p> <p>I recommend that you first construct the integer matrix M which will compute the (i+1,..., i+k) terms in your sequence based on the (i, ..., i+k-1) terms. (only row k+1 of this matrix is interesting). Compute the first k entries "by hand", then calculate M^(10^15) based on the repeated squaring trick, and apply it to terms (0...k-1). </p> <p>The (integer) entries of the matrix will grow exponentially, perhaps too fast to handle. If this is the case, do the very same calculation, but modulo p, for several moderate-sized prime numbers p. This will allow you to obtain your answer modulo p, for various p, without using a matrix of bigints. After using enough primes so that you know their product is larger than your answer, you can use the so-called "Chinese remainder theorem" to recover your answer from your mod-p answers.</p>
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    1. POProgramming problem - Game of Blocks
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    1. COFibonacci numbers mod P can be calculated using the matrix form of Fibonacci in `O(Log(N))`, where `N = 10^15`. For `C > 2` the problem can be converted to matrix for and similar matrix exponentiation algorithm applied as you mentioned, but the size of the matrix is way too large even to make one iteration of multiplication. It can be as large as `750 x 750 - O(N^3)` for a single multiplication. It is also known that the answer needs to be calculated `mod 100000007 (prime number)`, hence Chinese remainder theorem is not applicable.
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    2. COI see. I suggest you include the mod 100000007 requirement in the problem statement above.
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