Tobby_And_Friends's blog

By Tobby_And_Friends, history, 20 months ago, In English,

Link: https://uva.onlinejudge.org/external/108/10883.pdf

I could figure out that let's say for 5 numbers the answer is (Arr[1] + 4 * Arr[2] + 6 * Arr[3] + 4 * Arr[4] + Arr[5])/2^16. So the problem basically reduces to finding the values in the nth row of pascal's triangle and power of 2. But I am not very sure as to how to implement the solution. Any help is really appreciated.

 
 
 
 

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20 months ago, # |
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HINT: If you calculate in the paper, you will see answer is

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20 months ago, # |
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You can use the fact that , i.e. no need of pre-computation. You can find a similar problem here

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    20 months ago, # ^ |
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    Yeah, recursively like C(n,  k)  =  C(n,  k  -  1)  *  (n  -  k  -  1)  /  k

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20 months ago, # |
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HINT 2: Number of combinations can calculated recursively: C(n, k) = C(n, k - 1) * (n - k - 1) / k

Also C(n, 0) = 1

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20 months ago, # |
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HINT 3: For bigger n finding combination will not be work, so use logarithm : )

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20 months ago, # |
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If you couldn't solve, see solution with logarithm and numerical combinations