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Any function f(x) that is infinitely differentiable in the neighborhood of the point x = 0 can be expanded into a Maclaurin series
or more compactly,
Under certain conditions, f(x) is equal to its Maclaurin series for all x sufficiently close to 0.
Often, the Maclaurin series of f(x) is written up to only order xn-1 with a Peano remainder term, i.e.,
which facilitates the evaluation of f(x). The series being truncated to the xn-1 term, details about higher-order terms are inevitably lost. However, certain information concerning the series remains well-preserved. For instance, the Maclaurin series of 1/f(x) with an identical Peano remainder term can still be found.
Given the Maclaurin series of a function f(x) with a Peano remainder term, determine that of the reciprocal of f(x).
The input consists of a single test case.
The first line contains an integer n (1 ≤ n ≤ 100). Each of the next n lines contains a fraction pk/qk (0 ≤ k ≤ n - 1) where pk and qk (-100 ≤ pk ≤ 100, 1 ≤ qk ≤ 100) are integers coprime to each other in the form either pk/qk (if qk ≠ 1) or pk (if qk = 1). It is guaranteed that p0 ≠ 0.
The given numbers represent the Maclaurin series
of some function f(x).
Let the Maclaurin series of 1/f(x) be
where qk' > 0, and pk' and qk' are integers coprime to each other. Output n lines each of the form either pk'/qk' (if qk' ≠ 1) or pk' (if qk' = 1), listing the coefficients of 1, x, x2, ... , xn-1 in order.
10 1/2 -46/3 9/40 61/80 -22/5 -87/41 1/11 -42/5 -4/13 -21/44
2 184/3 169199/90 6223213/108 28611844561/16200 107867589093163/1992600 545525329585163771/328779000 401297234058997162673/7890696000 2398505849871035056778279/1538685720000 882189800570641237196128639/18464228640000
Case time limit(单组数据时间限制) | 2000MS |
时间上限 | 内存上限 |
5000 | 131072 |