case in point中文版

"Case in point"是一个英语短语，意为"典型例子"或者"明确事例"。它通常用于强调或者证明某个观点或者论点。这个短语可以用来引出一个特定的案例或者事件，以证明某种情况确实发生或者存在。

举个例子，假设我们在讨论环境污染对人类健康的影响。我们可以说，"大城市中的空气污染是一个很严重的问题，'case in point'就是北京市。北京市污染水平极高，每年都面临着雾霾的困扰。这个事例可以说明环境污染对人类健康的直接影响。"

在另一个例子中，我们在谈论教育的重要性。我们可能会说，"教育对个人发展至关重要，'case in point'就是某某名人。他出身贫寒，但通过努力学习取得了巨大的成功。这个例子表明了教育的力量和影响。"

总之，"case in point"在英语中用来指出特定的案例或者事件，以证明某一观点或者论点。它帮助我们以一个典型的例子来支持我们的说法或者论证，使得我们的观点更加具有说服力。

相关问题

public void onAccessibilityEvent(AccessibilityEvent event) { int eventType = event.getEventType(); switch (eventType) { case AccessibilityEvent.TYPE_VIEW_CLICKED: Point point = new Point(event.getX(), event.getY()); break; }，这段代码出现错误：Cannot resolve method 'getX' in 'AccessibilityEvent'是什么原因

AccessibilityEvent类中并没有getX()方法，getX()是View类中的方法。如果你想获取事件发生的坐标，可以使用AccessibilityEvent的getParcelableData()方法获取一个Parcelable类型的数据，然后将其转换为AccessibilityClickableSpanInfo对象，通过该对象的getX()和getY()方法获取坐标。具体代码如下：

public void onAccessibilityEvent(AccessibilityEvent event) {
    int eventType = event.getEventType();
    switch (eventType) {
        case AccessibilityEvent.TYPE_VIEW_CLICKED:
            Parcelable data = event.getParcelableData();
            if (data instanceof AccessibilityClickableSpanInfo) {
                AccessibilityClickableSpanInfo clickableSpanInfo = (AccessibilityClickableSpanInfo) data;
                Point point = new Point(clickableSpanInfo.getX(), clickableSpanInfo.getY());
                // do something with the point
            }
            break;
    }
}

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：