从0对数区间到第9对数区间,革螨种类数显示逐渐下降的趋势。
From log interval 0 to log interval 9, the mite species number shows a gradual descending.
我们则跳过比猜想数小的那个区间,然后我们重复这一过程,跟之前我们讲过的,递归思想非常类似,我们解决问题的时候,先把问题一步步变小,然后解决小问题。
So this is very similar, this is a kind of recursive thinking we talked about earlier, where we take our problem and we make it smaller we solve a smaller problem, et cetera.
其中荣誉值按间隔100做了划分(即0 - 99 100 - 199等等),而在每个区间上,取那个区间上的用户所作的回答投票的平均数。
I divided reputation ranges into blocks of 100 (i.e. 0-99, 100-199, etc.) and averaged the number of times users in that range voted up an answer.
So this is very similar, this is a kind of recursive thinking we talked about earlier, where we take our problem and we make it smaller we solve a smaller problem, et cetera.
我们则跳过比猜想数小的那个区间,然后我们重复这一过程,跟之前我们讲过的,递归思想非常类似,我们解决问题的时候,先把问题一步步变小,然后解决小问题。
And when I do this test, what I want to do, is say I'm going to pick the middle spot, and depending on the test, if I know it's in the upper half, I'm going to set my start at the mid point and the end stays the same, if it's in the front half I'm going to keep the front the same and I'm going to change the endpoint.
如果我知道目标数可能,再比中值点大的区间里,我可能就会把开始点设为中值点,而尾点不变,如果在小的那个区间里,就保持开始点不变而把尾点设为中值点,你们可以看到这儿的代码,就是这么做的,对不对?它是怎么做的?
And that idea was, we make a guess in the middle, we test it so this is kind of a guess and check, and if the answer was too big, then we knew that we should be looking over here. If it was too small, we knew we should be looking over here, and then we would repeat.
这些有理数是有序排列的,然后我们的想法是,首先在中间取个数作为猜想数,然后对这个猜想数进行验证,如果由猜想数得到的答案太大,我们知道应该跳过,比猜想数大的那个区间,如果太小的话。
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