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вернуться в форумThe answer Послано Li, Yi 17 авг 2001 09:26 n
(A[n+1] - A[0] - 2 * Sigma(k * c[n - k + 1]))
k=1
A[1] = A[0] + ----------------------------------------------
n + 1 Re: The answer Послано VladG 22 окт 2002 22:09 How, did you discover this expression?
> n
> (A[n+1] - A[0] - 2 * Sigma(k * c[n - k + 1]))
> k=1
> A[1] = A[0] + ----------------------------------------------
> n + 1 Re: The answer Послано Antikr 4 май 2007 02:45 ........................................................... n
..................... (A[n+1] - A[0] - 2 * Sigma(k * c[n - k + 1]))
......................................................... k=1
A[1] = A[0] + ---------------------------------------------
...................................................... n + 1
did you mean this??
Edited by author 04.05.2007 02:46 Re: The answer In this problem, you can also use simply binary search, note that :
A[i] = 2*(A[i-1]+C[i])-A[i-2]; |
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