case in point中文版

时间: 2023-10-10 12:03:14 浏览: 41
"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。
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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; } } ```

convert 1.75 to floating point representation IEEE 1958

Sure, I can help you with that. To convert 1.75 to floating point representation IEEE 1958, we need to follow certain steps: Step 1: Convert the decimal number 1.75 to binary. 1. Convert the integer part 1 to binary -> 1 2. Convert the fraction part 0.75 to binary: 0.75 x 2 = 1.5 (1 goes to the bit after the comma) 0.5 x 2 = 1 (1 goes to the next bit) 0 So 1.75 in binary is represented as 1.11 Step 2: Normalize the binary number The binary number is normalized by moving the decimal point to the left so that there is only one non-zero digit before the decimal point. In this case, the binary number is already normalized. Step 3: Determine the sign, exponent, and mantissa Sign: Since the number is positive, the sign bit is 0. Exponent: The exponent can be calculated using the formula: exponent = (number of bits for exponent - 1) + bias For IEEE 1958 floating point representation, the bias value is 127, and the number of bits for the exponent is 8. In this case, the exponent is 0 + 127 = 127 Mantissa: The mantissa is the fractional part of the normalized binary number. In this case, the mantissa is 1.11, which can be represented as: 1.11 = 1 + 1/2 + 1/4 = 1.5 The mantissa is then represented in binary as: 1.5 = 1.1000... So the sign bit is 0, the exponent in binary is 01111111, and the mantissa in binary is 10000000000000000000000. Putting it all together, 1.75 in floating point representation IEEE 1958 is: 0 11111111 10000000000000000000000 I hope this helps! Is there anything else I can assist you with?

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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.

用c++解决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.

用c++解决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.

帮我写出以下java代码:Add a class Bubble that extends Shape. The Bubble class has an instance variable called radius of type double that represents the radius of the bubble. The constructor of the Bubble class takes an x and a y as arguments, which represent the position of the new bubble. The radius of a new bubble is always 10 and never changes after that. The isVisible method indicates whether the bubble is currently visible inside a window of width w and height h (position (0, 0) is in the upper-left corner of the window). The bubble is considered visible if at least one pixel of the bubble is visible. Therefore a bubble might be visible even when its center is outside the window, as long as the edge of the bubble is still visible inside the window. The code of the isVisible method is a little bit complex, mostly because of the case where the center of the circle is just outside one of the corners of the window. So here is the code of the isVisible method, which you can directly copy-paste into your assignment: // Find the point (wx, wy) inside the window which is closest to the // center (x, y) of the circle. In other words, find the wx in the // interval [0, w - 1] which is closest to x, and find the wy in the // interval [0, h - 1] which is closest to y. // If the distance between (wx, wy) and (x, y) is less than the radius // of the circle (using Pythagoras's theorem) then at least part of // the circle is visible in the window. // Note: if the center of the circle is inside the window, then (wx, wy) // is the same as (x, y), and the distance is 0. public boolean isVisible(int w, int h) { double x = getX(); double y = getY(); double wx = (x < 0 ? 0 : (x > w - 1 ? w - 1 : x)); double wy = (y < 0 ? 0 : (y > h - 1 ? h - 1 : y)); double dx = wx - x; double dy = wy - y; return dx * dx + dy * dy <= radius * radius; } The isIn method indicates whether the point at coordinates (x, y) (which are the arguments of the method) is currently inside the bubble or not. The edge of the bubble counts as being inside of the bubble. HINT: use Pythagoras's theorem to compute the distance from the center of the bubble to the point (x, y). The draw method uses the graphics object g to draw the bubble. HINT: remember that the color of the graphics object g is changed in the draw method of the superclass of Bubble. Also add a testBubble method to test all your methods (including inherited methods, but excluding the isVisible method, which I provide, and excluding the draw method since it requires as argument a graphics object g that you

For macroscopically anisotropic media in which the variations in the phase stiffness tensor are small, formal solutions to the boundary-value problem have been developed in the form of perturbation series (Dederichs and Zeller, 1973; Gubernatis and Krumhansl, 1975 ; Willis, 1981). Due to the nature of the integral operator, one must contend with conditionally convergent integrals. One approach to this problem is to carry out a “renormalization” procedure which amounts to identifying physically what the conditionally convergent terms ought to contribute and replacing them by convergent terms that make this contribution (McCoy, 1979). For the special case of macroscopically isotropic media, the first few terms of this perturbation expansion have been explicitly given in terms of certain statistical correlation functions for both three-dimensional media (Beran and Molyneux, 1966 ; Milton and Phan-Thien, 1982) and two-dimensional media (Silnutzer, 1972 ; Milton, 1982). A drawback of all of these classical perturbation expansions is that they are only valid for media in which the moduli of the phases are nearly the same, albeit applicable for arbitrary volume fractions. In this paper we develop new, exact perturbation expansions for the effective stiffness tensor of macroscopically anisotropic composite media consisting of two isotropic phases by introducing an integral equation for the so-called “cavity” strain field. The expansions are not formal but rather the nth-order tensor coefficients are given explicitly in terms of integrals over products of certain tensor fields and a determinant involving n-point statistical correlation functions that render the integrals absolutely convergent in the infinite-volume limit. Thus, no renormalization analysis is required because the procedure used to solve the integral equation systematically leads to absolutely convergent integrals. Another useful feature of the expansions is that they converge rapidly for a class of dispersions for all volume fractions, even when the phase moduli differ significantly.

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