uva 766 sum of power
題意: 等冪求和 輸入k次冪,化簡成,按M,ak+1,ak, ... , a0 的順序輸出
伯努利數Bn是等冪求和的解析解中最為明顯的特徵,定義等冪和如下,其中m, n ≥ 0:
- {\displaystyle S_{m}(n)=\sum _{k=1}^{n}k^{m}=1^{m}+2^{m}+\cdots
+{n}^{m}}
這數列和的公式必定是變數為n,次數為m +1次的多項式,稱為伯努利多項式。伯努利多項式的係數與伯努利數有密切關係如下:
- {\displaystyle S_{m}(n)={\frac {1}{m+1}}\sum
_{k=0}^{m}{\binom {m+1}{k}}B_{k}^{+}n^{m+1-k},}
其中(m +
1
k) 為二項式係數。
伯努利數的遞推式:
即B0 = 1
由於k很小,把需要的組合數,和伯努利數打表即可。( 我偷懶了。。。 給個提示。。。伯努利數前二十項維基百科有的。。)
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uva 766 sum of power
題意: 等冪求和 輸入k次冪,化簡成,按M,ak+1,ak, ... , a0 的順序輸出 伯努利數Bn是等冪求和的解析解中最為明顯的特徵,定義等冪和如下,其中m, n ≥ 0: {\displaystyle S_{m}(n)=\sum _{k=1}^{n}k^{m
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