另一个按位运算符拼图 [英] Another bitwise operators puzzle
问题描述
给定一个数字，我们需要将其除以3，而不使用*，/，％运算符。那么问题还提到itoa（）函数可用!!
Given a number, we need to divide it by 3, without using *, /, % operators. Well the question also mentions that itoa() function is available !!
推荐答案
给定一个数字，我们需要除以3，不使用*，/，％运算符。那么问题还提到itoa（）函数可用!!
Given a number, we need to divide it by 3, without using *, /, % operators. Well the question also mentions that itoa() function is available !!
我没有尝试过itoa但它有点像跟随，希望它有所帮助：
I have not tried with itoa but it is something like following, hope it helps:
zodilla58'的解决方案将起作用，我想。但是，如果你已经达到分子，则通过除数检查向上计数必须称为强力算法：可用于小值，或者如果你只想做一次。
让我们看看我们是否不能提出更具可扩展性的东西。
提到itoa，表明我们可能想看一下10数字。
让我们看看3：27,33,102,300,402的几个倍数
啊！
看起来如果你将基数10加在一起，那么当有3的精确倍数时它们将总和为3,6或9。
这是真的吗？谁能找到一个反例？任何人都可以提供这样的数学证明吗？
编写一个快速程序可能更容易测试是否在3的第一个百万倍数上是真的。证明？也许没有，但我会相信。
假设它是真的，算法是直截了当的。
- 转换为ascii字符在基地10（itoa）
- 提取单个字符
- 将每个字符转换为小整数（0：9）
- 和整数
- 如果总和> 9，重复。
- 比较最终的，单个数字的总和与3,6和& 9
zodilla58''s solution will work, I think. However, counting upwards by the divisor checking all the time if you have reached the numerator must be called a brute force algorithm: workable for small values, or if you only want to do this once.
Let''s see if we can''t come up with something a little more scalable.
The mention of itoa, suggests that we might want to look at the base 10 digits.
Let''s look at a few multiples of 3: 27, 33, 102, 300, 402
Ah!
It looks like if you add the base 10 digits together, then they will sum to 3, 6 or 9 when there is an exact multiple of 3.
Is this true? Can anyone find a counter-example? Can anyone provide a mathematical proof that this has to be so?
It might be easier to write a quick program to test if this is true on the first million multiples of 3. Would that be proof? Maybe not, but I would be convinced.
Assuming it is true, the algorithm is straight-forward.
- convert to ascii character in base 10 ( itoa )
- extract individual characters
- convert each character to a small integer ( 0:9 )
- sum integers
- if sum > 9, repeat.
- compare final, single digit sum with 3, 6 & 9
看起来如果你将基数10位加在一起，那么当有一个数字时，它们总和为3,6或9 3的确切倍数。
这是真的吗？谁能找到一个反例？任何人都可以提供这样的数学证明吗？
It looks like if you add the base 10 digits together, then they will sum to 3, 6 or 9 when there is an exact multiple of 3.
Is this true? Can anyone find a counter-example? Can anyone provide a mathematical proof that this has to be so?
这显然不是3333的情况。这可以除以3，但总和为12.除非你再次加12 ...然后你会得到3而你是正确的。但是，如果没有，你有一个反例。
现在，如果我记得正确的表格学校你可以检查一个数字是否可以除以3，如果基数10位数的总和返回一个可除以3的数字。
我没有这方面的证据，这只是他们在数学上给我们买的东西:)
希望这会有所帮助
欢呼，
stefaan
this is obviously not the case take 3333. this is divisable by 3 but sums to 12. unless you sum 12 again ... then you would get 3 and you are correct. however if not there you have a counter example.
Now if I remember correct form school you can check if a number is divisable by 3 if the sum of the base 10 digits returns a number divisable by 3.
I don''t have proof for this, it''s just something they tought us at math :)
hope this helps
cheers,
stefaan
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