共有8个硬币:
页: 1
It is possible to make $2 in the following way: 1x$1 + 1x50p + 2x20p + 1x5p + 1x2p + 3x1p
How many different ways can $3 be made using any number of coins?
我们如何使用购买力平价?
共有8个硬币:
页: 1
It is possible to make $2 in the following way: 1x$1 + 1x50p + 2x20p + 1x5p + 1x2p + 3x1p
How many different ways can $3 be made using any number of coins?
我们如何使用购买力平价?
这个问题和类似问题最好通过解决。
为此,请考虑形式条件。
1/(1 - x^k) = 1 + x^k + x^(2k) + x^(3k) + ...
<代码>k为:1p、2p、5p、10p、20p、50p、100p、200p
。 如今,这8条术语全部齐头并进,以便共同处理。
f(x) = 1 / [ (1-x) * (1-x^2) * (1-x^5) * ... * (1-x^200) ]
然后,<代码>x^m的系数就是指从特定名称中提取的<代码>>>/mp/code>的序号。 例如,<代码>x^200>的系数为6
,与从特定名称中获取<代码>200p = 2的准确方式相符。
这里对这项工作的原因作了迅速和 d脏的解释。 <代码>x^m> >
i1*k1 + i2*k2 + ... + i8*k8 = m
现在,<代码>k1 = 1对应的术语为1p
,“代码>k2 = 2对应的术语为2p
,而“代码>k3=5p对应的术语为<代码>5p,等等。 以上数额改为
i1*(1p) + i2*(2p) + i3*(5p) + ... + i8*(200p) = m
规定每个教派的占有额。
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