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是会计专业? 整页都在说钱
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不会解唉
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Q3: 8, 16, 32, 64
Q4: 1, 8, 16, 57
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求步骤...谢谢
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4年级的作业
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Current coins are 2 austral, 4 austral, 25 austral. And the coins in any amount must be unique.
2, 4, 25
1. Let's say if the amount is 7 australs.
Existing coins (2+4=6) would not be able to cover. So 1 austral coin needs to be issued.
1, 2, 4, 25
2. Let's say if the amount is 8 australs. Existing coins (1+2+4=7) would not be able to cover without repeating. So will need to issue 8 austral coin.
1, 2, 4, 8, 25
3. Let's say if the amount is 16 australs. Existing coins (1+2+4+8=15) would not be able to cover without repeating. So will need to issue 16 austral coin.
1, 2, 4, 8,16, 25
4. Let's say if the amount is 57 australs. Existing coins (1+2+4+8+16+25=56) would not be able to cover without repeating. So will need to issue 57 austral coin.
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第三题和第四题一样的:七个硬币面值应该是1,2,4,7,13,25,50。
解题方法是两分法:
一:100/2=50,剩下的六个硬币加起来能够达到1-50所有整数就行;
二:50/2=25, 剩下的五个硬币加起来能够达到1-25所有整数就行;
三:25/2=12.5 ,四舍五入 13,剩下四个硬币加起来能够达到1-12所有整数就行
四:13/2=6.5 ,四舍五入 7,剩下三个硬币加起来能够达到1-6所有整数就行
五:7/2=3.5 ,四舍五入 4 ,剩下两个硬币加起来能够达到1-3所有整数就行
六:4/2=2
七:2/2=1
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Q3 : we have 1, 2, 4
we can make 1-7
we need a 8
then we can make 8-15
after that we need a 16
so we can make 16-31,by using 1,2,4,8,16
we can not make 32 by using current coins, so we need a 32
the same way we can make 32-63
the last coin, we need a 64 ,so we can make 1-100, without using them more than once.
Q4: we have 2, 4, 25
we need 1, then 8 ,then 16
with these 6 coins ,we can make up to 56,so we need a 57 to cover the rest from 57 to 100.
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Q3 ,8.16,32,64
Q4 1,8,16,31,57
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其实这两道题有多个解的。
1,2,4,7,13,25,50七个硬币就可以。
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其实,这道题在大学学数学的时候,会再遇到,那时候就抽象多了。
构建一个整数集,可以产生不同的子集和。
http://www.combinatorics.org/Volume_5/PDF/v5i1r3.pdf
http://www.openproblemgarden.org ... istinct_subset_sums
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Terry 的数学,不是应该很好吗?还出高中数学书呢。
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