For example, suppose we have four solutions in Q as shown in Table 2. Each solution has a label and a membership degree with respect to the positive class predicted by the classifier. When the accuracy of the classifier is larger than or equal to 70%, these solutions are ranked in descending order with respect to their membership degrees (i.e., s2 > s3 > s1 > s4). When the accuracy of the classifier is larger than or equal to 30% and smaller than 70%, the positive solutions are ranked in ascending order with respect to their membership degrees (i.e., s3 > s2). Next, the negative solutions are ranked in descending order with respect to their membership degrees (i.e., s1 > s4). Then, the positive solutions are ranked before the negative solutions (i.e., s3 > s2 > s1 > s4). When the accuracy of the classifier is smaller than 30%, these solutions are ranked in ascending order with respect to their membership degrees (i.e., s4 > s1 > s3 > s2).

这段文字描述了一个在分类器准确率不同范围内如何对解决方案进行排名的过程。其中，每个解决方案都有一个标签和一个属于正类的隶属度。当分类器准确率大于等于70%时，这些解决方案按照隶属度从大到小排名。当分类器准确率大于等于30%且小于70%时，正类解决方案按照隶属度从小到大排名，负类解决方案按照隶属度从大到小排名。然后，正类解决方案排在负类解决方案前面。当分类器准确率小于30%时，这些解决方案按照隶属度从小到大排名。

相关问题

分 Suppose the hard disk with SCAN scheduling Alg has a single platter （盘面） as illustrated below , and the rotation speed for the platter is 6000 rounds per minute . The disk has only 200 tracks , and each track is composed of 100 sectors . The time cost for the disk arm to navigate between neighboring track is about 1ms. Suppose the disk arm currently hangs over the 100th track , and moves centrifugally towards lower - order tracks on the outside . The pending disk block requests are positioned on the tracks with numbers 50, 180,90,30,120, respectively , and form a queue . For each block in this queue , we do not know what are its exact sector numbers , and we can only assume that the corresponding sector randomly distributes on each track as shown below . Then , The expected time to finish the retrieval of the five sectors is1] ms . If SSTF is used , the expected time is [] ms . 0 磁道 磁头运动方向 .

根据题目所给的信息，我们可以画出磁道与扇区的示意图：

   |       Track 1        |
   | Sector 1 | Sector 2 |
   |---------------------|
   |       Track 2        |
   | Sector 1 | Sector 2 |
   |---------------------|
   |         ...          |
   |---------------------|
   |      Track 200       |
   | Sector 1 | Sector 2 |

其中，每个磁道有 100 个扇区，总共有 200 个磁道。

根据题目所描述的情况，我们可以计算出使用 SCAN 调度算法时，磁头移动的距离为：

- 从 100 磁道移动到 50 磁道，需要移动 50 个磁道，耗时 50ms；
- 从 50 磁道移动到 30 磁道，需要移动 20 个磁道，耗时 20ms；
- 从 30 磁道移动到 90 磁道，需要移动 60 个磁道，耗时 60ms；
- 从 90 磁道移动到 120 磁道，需要移动 30 个磁道，耗时 30ms；
- 从 120 磁道移动到 180 磁道，需要移动 60 个磁道，耗时 60ms。

因此，总共需要移动 220 个磁道，耗时为 220ms。

对于 SSTF 调度算法，我们可以按照距离当前磁头位置最近的磁道顺序来访问请求的磁道。因此，访问的顺序应该是 90、120、50、30、180。这样，总共需要移动的磁道数为 130，耗时为 130ms。

因此，答案为：

- 使用 SCAN 调度算法时，耗时为 220ms；
- 使用 SSTF 调度算法时，耗时为 130ms。

用c++解决Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points specializing in the same pair of currencies. Each point has its own exchange rates, exchange rate of A to B is the quantity of B you get for 1A. Also each exchange point has some commission, the sum you have to pay for your exchange operation. Commission is always collected in source currency. For example, if you want to exchange 100 US Dollars into Russian Rubles at the exchange point, where the exchange rate is 29.75, and the commission is 0.39 you will get (100 - 0.39) * 29.75 = 2963.3975RUR. You surely know that there are N different currencies you can deal with in our city. Let us assign unique integer number from 1 to N to each currency. Then each exchange point can be described with 6 numbers: integer A and B - numbers of currencies it exchanges, and real RAB, CAB, RBA and CBA - exchange rates and commissions when exchanging A to B and B to A respectively. Nick has some money in currency S and wonders if he can somehow, after some exchange operations, increase his capital. Of course, he wants to have his money in currency S in the end. Help him to answer this difficult question. Nick must always have non-negative sum of money while making his operations. Input The first line contains four numbers: N - the number of currencies, M - the number of exchange points, S - the number of currency Nick has and V - the quantity of currency units he has. The following M lines contain 6 numbers each - the description of the corresponding exchange point - in specified above order. Numbers are separated by one or more spaces. 1 ≤ S ≤ N ≤ 100, 1 ≤ M ≤ 100, V is real number, 0 ≤ V ≤ 103. For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10-2 ≤ rate ≤ 102, 0 ≤ commission ≤ 102. Let us call some sequence of the exchange operations simple if no exchange point is used more than once in this sequence. You may assume that ratio of the numeric values of the sums at the end and at the beginning of any simple sequence of the exchange operations will be less than 104. Output If Nick can increase his wealth, output YES, in other case output NO.

以下是用c++解决这道题目的思路：

首先，我们需要将输入存储在合适的数据结构中。因为每个货币对之间可能有多个交换点，所以我们可以使用一个二维数组来存储每个货币对的交换点信息。另外，我们可以使用一个一维数组来存储每个货币的当前价值。

接下来，我们可以使用深度优先搜索算法来遍历所有可能的交换序列。在搜索过程中，我们需要记录当前的货币和余额。对于每个交换点，我们需要计算出交换后的货币和余额，并将其作为新的状态进行搜索。

如果在搜索过程中找到一种交换序列使得余额增加，则输出YES，否则输出NO。

以下是c++的代码实现：