solution
动态规划训练赛1
竞赛时间：2017年7月21日13:00-16:00
题目名称邮局选址 补兵任务
输入文件名post.in cs.in mission.in 输出文件名post.out cs.out mission.out 每个测试点时限 1 sec 2 sec 1 sec
内存限制128M128M128M
测试点数目101010
每个测试点分值101010
是否有部分分无无无
题目类型传统传统传统提交源程序须加后缀
对于Pascal语言post.pas cs.pas mission.pas 对于C 语言post.c cs.c mission.c
对于C++ 语言post.cpp cs.cpp mission.cpp 注意：最终测试时，所有编译命令均不打开。评测系统为Win7。
邮局选址
【问题描述】
有n 个村庄分布在一条直线上，每个村庄可以用一个坐标i x 来进行描述。现
在，你需要建设m 个邮局，使得每个村庄到离它最近的邮局的距离之和最小。 【数据规模和约定】
对于%100的数据，满足3001≤≤n ，301≤≤m ，100001≤≤i x 。 【简要题解】
首先，如果只能放一个邮局，显然放在中位数最优。
如果放多个邮局，肯定是将村庄按顺序拆成若干段，每段取中位数。于是
可以进行DP ：设]][[j i f 表示到第i 个村庄已经放了j 个邮局的最小距离和。预处理出]][[j i g 表示从i 到j 的村庄用一个邮局的最小值后，可以列出方程：
f i j =min {f k ?1j ?1+
g k i }
【问题描述】
对于一名DotA玩家，补兵的个数(creep score)是衡量一名选手的能力的一项重要指标。特别是在打路人局的时候，补兵的能力就更加关键，因为常常会有队友和你抢补刀。比如，队友操控的老鹿开着大在收一波兵，你操作的英雄如果是幽鬼，就需要在老鹿的AOE中偷偷补上几刀来保证自己的发育。
我们可以建立如下模型来大致模拟这种状况：现在有n个小兵，每个小兵都有自己的血量（血量一定是正整数）。你和老鹿轮流对小兵进行攻击。每次，你可以选择对某个小兵造成1点伤害（或者你可以选择不造成任何伤害）；接着，老鹿会对所有小兵都造成1点伤害。如此往复，直到所有小兵都死亡（血量达到0）。在这一过程中，如果你对某个小兵造成了致命伤害（使它的血量由1变为0），那么你就算成功补到这个兵。
对于给定的情形，你需要计算你的最大补兵数。
【数据规模和约定】
对于30%的数据，满足n≤100。
对于50%的数据，满足n≤500，T≤30，每个小兵的血量不超过500。
对于100%的数据，满足1≤n≤1000，1≤T≤70。
【简要题解】
考虑一个最特殊的例子：小兵的血量分布恰好长成1,2,3,….。如果是这样，我们显然可以补到每一刀。
现在的状况是，我们面对的局面并不是这样，所以我们应该先攻击一些小兵，使得小兵的血量分布变成上面描述的情况（有一个专门的术语就描述了这种操作，叫做“垫刀”）。
还可以明确的情况是，如果有两个小兵的血量一直是一样的，那么这两个小兵我们至多只能补到一个。所以，如果我们发现在血量i没有别的小兵，应该从血量高于i的小兵中，挑一个血量有和别的小兵相同的、血量尽量少的来补齐。这样，我们可以当作这个小兵的血量就是i，只是在选择它前需要垫几刀。
具体到DP，设f I,J表示考虑到血量为i的小兵，在操作的过程中省下了j刀的情况下所能补到的最大兵数。如果第i个小兵不补，则f I,J=f IKL,JKL；如果要补，则f I,J=f IKL,J＋M[I]+1。其中c[i]表示要补血量被当作i的小兵之前要垫多少刀。两者取一个最大值即可。转移的一些细节可以参考标程。
【问题描述】
n 个人正在完成m 件任务。
每个人都有一定的能力，我们可以用正整数a R 来描述每个人的能力；每件任务都有一定的难度，我们可以用正整数b T 来描述每个任务的难度。具体地，编号为i 的人完成编号为j 的任务所需要的时间为a R ×b T 。
现在，你需要让n 个人按编号依次完成m 件任务，即第一个人从第一件任务开始，一直完成到最后一件；当第一个人完成第一件任务时，第二个人开始进行第一个任务……在此过程中，你需要保证一个任务只能正在由一个人完成。 你需要求出最后一个人完成最后一件任务的最早时间。 【数据规模和约定】
对于100%的数据，满足1≤n,m ≤100000，1≤a I ,b J ≤10000。 【简要题解】
根据题意，我们可以设i F 表示第i 个人的最早开始时间，其中01=F 。另外，我们设∑==j
k k j b S 1。则我们容易得出以下DP 式：
}max{111j i j i i i S a S a F F ×?×=?+++
显然直接DP 是过不去的，必须进行优化：
考虑构造向量),(1j j S S P ?=+，向量),(1+=i i a a Q 。则我们只需要求向量Q 和
所有的向量的点积最大值。显然这在凸壳上而且单峰，三分即可。
HW2-Solutions
PART A: Minimum Inventory Production Plan (i.e., one that allows arbitrary hiring and firing) A B C D E F G H I J K Forecast Net Cummulative Minimum # # Units Cummulative Inv. above Actual Demand Demand Net Demand of Workers Produced Production Min required Inventory C/32 E x 32 I - D July 1250 1050 1050 33 2 0 1056 1056 6 306 Aug 1100 1100 2150 35 0 2 1120 2176 26 326Sep 950 950 3100 30 5 0 960 3136 36 336 Oct 900 900 4000 29 1 0 928 4064 64 364 Nov 1000 1000 5000 32 0 3 1024 5088 88 388 Dec 1150 1250 6250 40 0 8 1280 6368 118 518 8 13 338 2238A B C D E F G H I J K Forecast Net Cummulative Minimum # # Units Cummulative Inv. above Actual Demand Demand Net Demand of Workers Produced Production Min required Inventory C/32 E x 32 I - D July 1250 1050 1050 33 2 0 1056 1056 6 306 Aug 1100 1100 2150 35 0 2 1120 2176 26 326 Sep 950 950 3100 29 6 0 928 3104 4 304 Oct 900 900 4000 28 1 0 896 4000 0 300 Nov 1000 1000 5000 32 0 4 1024 5024 24 324 Dec 1150 1250 6250 39 0 7 1248 6272 22 422 9 13 82 1982A B C D E G H I J Forecast Net Cummulative Cum. Units Monthly Cummulative Inv. Above Actual Demand Demand Net Demand produced Production Production Min required Inventory per worker 32 x 34 H - D July 1250 1050 1050 32 1088 1088 38 338Aug 1100 1100 2150 64 1088 2176 26 326Sep 950 950 3100 96 1088 3264 164 464Oct 900 900 4000 128 1088 4352 352 652Nov 1000 1000 5000 160 1088 5440 440 740Dec 1150 1250 6250 192 1088 6528 278 6781298 3198PROBLEM # 1 Month Fire Hire Minimum inventory required per month = 300 Ending Inventory = 400 Employee-hours required per hour = 5 hours Working days per months = 20 Each employee works 8-hour-day Workforce at the end of June = 35K CALCULATION K (Units produced by one worker in one day)Initial Inventory = 500 TOTAL K = 8/5 = 1.6# Units Produced by 1 worker in a month = 1.6 x 20 = 32 In this problem there is sufficient inventory on hand to reduce the workforce by one worker in the months of Sep, Oct and Dec. That decision will allow to reduce the total inventory above minimum required from 338 units to 82. Month F Month Fire Hire TOTAL D/E PART B: Production Plan that meets demand but does not hire or fire workers during the six-month period TOTAL 32 32 33 Because there are 35 workers employed at the end of June, the constant workforce plan require fire 1 worker at the beginning of July. No further hiring or firing of workers are required. The inventory levels that results from a constant workforce of 34 workers appear in the following table. 33 Ratio 33 34
McKinseySolutions_OHI
Organizational Health Index (OHI) Using ‘Organizational Health’ to build and sustain financial performance
Linking organizational health and performance Our research has proven that the healthiest companies are more than twice as likely to have EBITDA margins above their industry median Shifting your organization toward higher performance u Assess underlying mindsets and behaviors that impact financial performance u Benchmark your organization’s health index against your industry and high performers u Understand your organization’s archetype to identify the six to seven “must-have” practices for success u Prioritize intervention programs needed to improve health u Periodically track health to ensure ongoing progress and catch potential issues early Connecting with experts u Gain important, and often unexpected, insights identified by analysts with experience across many companies, industries, and geographies u Develop the best solution for your company by drawing upon McKinsey’s deep organizational expertise and broad experience serving 96% of the world’s leading global institutions About OHI The Organizational Health Index solution measures and tracks the organizational elements that drive financial performance. It provides a simple but powerful road map for leaders and managers to improve organizational health. “McKinsey’s behavioral work is the most profound I’ve been involved in – the impact is extraordinary.” Chief Executive Officer,Global Telecoms Company Nine elements of organizational health Proven link to financial performance
solutions manual for Fundamentals of Corporate
solutions manual for Fundamentals of Corporate Finance 9th edition by Ross Source: https://www.wendangku.net/doc/093089476.html,/Archive/Cobol/https://www.wendangku.net/doc/093089476.html,ng.cobol/2010?03/msg00035.html ? From: Robert Carter
Filter_Solutions_教程
用Filter solutions设计滤波器笔记 要求：分别设计5阶切比雪夫1型带通滤波器、切比雪夫2型带通滤波器，中心频率500MHz，带宽50MHz。尝试选用不同的电路结构，观察所用元器件的值有何变化。 切比雪夫1型滤波器：通带中等纹波，阻带是单调的 切比雪夫2型滤波器：通带中单调，阻带等纹波，即阻带中有零点（陷波点）。 （1）5阶切比雪夫1型带通滤波器：通带中等纹波，阻带是单调的 图1：参数设置 注释：Asymmetrical非对称的，只有带通滤波器才有此选项，用于设计非对称结构的滤波器。 Delay Equalize延迟补偿，低通和带通滤波器才有此选项，用于低通滤波器或者带通滤波器的通带群延迟补偿。延迟补偿的概念： Tx Line:Transmission Line 传输线滤波器 Sw Cap:Switched Capacitor开关电容滤波器， Active：有源滤波器， Passive:无源滤波器， Digital:数字滤波器， 1st Ele Shunt:滤波器电路结构的第一个元件是并联电感。 1st Ele Series:滤波器电路的第一个元件是串联电容。 Incl Source Bias：Check to include the bias due to the source resistance to appear in the transfer function, frequency response, impedance response, and time response of your filter. For example, if the source resistance is equal to the load resistance, checking this box will cause a -6dB offset will appear in the frequency response, and a factor of .5 will appear in the transfer function and time response. This offset goes away when this box is left unchecked. This box should be left unchecked when using the transfer function for any purpose other than passive filters applications. Complex Terminations:创建滤波器复杂的驱动终端，即自定义滤波器的源终端。 Add Stop Band Zeros：添加阻带零点，Butterworth and Chebyshev I 才有添加阻带零点的选项，这种滤波器已经不再是严格意义上的Butterworth and Chebyshev 滤波器。
雅思大作文范文：学习与打工 Causes and solutions
范文 Nowadays, there are a lot of students who work while studying. Although this tendency may lead to negative results, the number of young people who sacrifice education for work is growing. I believe that this problem has certain causes and can be solved by taking special measures. 现在，有很多学生一边学习一边工作。尽管这种趋势可能导致消极的结果，但是牺牲教育来换取工作的年轻人的数量正在增加。我认为这个问题有一定的原因，可以通过采取特殊措施来解决。 It is apparent that most of the students who work have many financial expenses to meet. One common cause is high cost of education. As many colleges and universities set high tuition fees, some families cannot fully afford the higher education for their children. Consequently, these students have to work to pay university and college fees. The other cause is living expenses. Many students study away from hometown, and have to pay for accommodation, food, entertainment etc. As it is often hard to cover these expenses, students are forced to earn money to afford their living. I think that the problem of students working during their studies results in lower quality of education and has no benefits at all. Thus, it should be solved on the governmental level. 很明显，大多数学生的工作有很多财政支出需要满足。一个共同的原因是高教育成本。由于许多学院和大学设置了很高的学费，一些家庭无法完全负担他们的孩子的高等教育。因此，这些学生必须工作来支付大学和学院的费用。另一个原因是生活费用。许多学生在外地学习，不得不支付住宿、食物、娱乐等费用。由于这些费用经常难以支付，学生被迫挣钱来维持生活。我认为学生在学习期间工作的问题导致了教育质量的下降，没有任何好处。因此，它应该在政府层面得到解决。 I can suggest two possible solutions to this problem. Firstly, the government could make higher education free. For instance, this result can be achieved by financing educational establishments from the country’s budget. Not only will it make universities and colleges accessible for everyone, but it will also reduce the number of working students. The second solution is promoting unpaid e-learning. Such form of distance education doesn’t require a lot of resources to be maintained. What’s more, students don’t have to leave their homes and can plan their schedules the way they prefer. 对于这个问题，我可以提出两种可能的解决办法。首先，政府可以免费提供高等教育。例如，这一结果可以通过从国家预算中资助教育机构来实现。它不仅将使大学和学院对每个人开放，而且还将减少就业学生的数量。第二个解决方案是推广无薪电子学习。这种形式的远程教育不需要大量的资源来维持。更重要的是，学生不必离开他们的家，可以计划他们的时间表，他们喜欢的方式。 In conclusion, I believe that mainly lack of financial resources causes students to work, but this problem can be solved by lowering the amount of financial expenses students have to meet.
Sample problems Solutions_2017
Sample Midterm for Risk Management: Applications to Financial Firms Multiple Choice 1.No matter what happens, a call option can never be worth more than the stock. a.True b. False c.Not if dividend yield is higher than the risk free rate Ans : True 2.DeadBeat stock has dropped from $100 to $3 in the last year, since its technology is now obsolete. A friend is offering to sell you an at the money American put with maturity of 5 years for a premium of $4. Do you: a.Grab this opportunity and buy the put b.Pass on thi s opportunity (and never speak to this ‘friend’ again) c.I’m not sure, since the question setup doesn’t talk about dividends at all Ans b. Since the price of the put can never be greater than the strike 3. A forward contract is usually traded a.Over the counter b.On the Chicago Mercantile exchange c.On the NYMEX Ans: a. 4.True or False. A forward is better than a future because it has lower credit risk. a.True b.False Ans: False 5.After a futures transaction the open interest: a.Increases b.Decreases c.Stays the same d.All of the above are possible And d. Quantitative Problems 1.The spot price of oil is $80 per barrel and the cost of storing a barrel of oil for one year is $3, payable at the end of the year. The risk-free interest rate is 5% per annum, continuously compounded. What is an upper bound for the one-year futures price of oil? The present value of the storage costs per barrel is 3e-0..05×1 = 2.854. An upper bound to the one- year futures price is (80+2.854)e0.05×1 = 87.10.
filter_solutions教程
让我们5分钟学会filter solutions速成教程。（由于本人在网上找了很久都没有找到关于Filter Solutions的教程，这几天接触了一些，就自己做一个简易教程和大家一起学习，由于本人水平有限，做得不好的请多包涵）。 一、首先了解一下滤波软件中的英文是什么意思。浏览一遍，待会不会的再回头查看。 滤波器软件英汉翻译及操作说明 (王Z G，湖南) Lowpass notch filters :低通陷波滤波器Order: 阶 filter circuits:滤波电路 frequency response:幅频响应Passband :通频带、传输带宽repeatedly cycle：重复周期maximum signal to noise ratio：最大信噪比 gain constants：增益系数，放大常数circuit topologies：电路拓扑结构 gain shortfall：增益不足 maximum output：最大输出功率 last stage：末级 preceding stage:前级 stage filter：分级过滤器 Gain Stage：增益级 voltage amplitude：电压振幅Component values: 元件值maximum valued: 最大值 minimum valued: 最小值 standard value:标准值 resistors: 电阻器 capacitors:电容器 operational amplifiers:运算放大器(OA) circuit board:(实验用)电路板 active filters:有源滤波器 supply currents:源电流 power supplies:电源 bypassing capacitors:旁路电容optimal:最佳的;最理想的 Gain Bandwidth:带宽增益 passive component:无源元件 active component: 有源元件 overall spread:全局;总范围Component characteristics:组件特性Modification:修改;更改 data book:数据手册 typical values:标准值;典型值 default values:省略补充 program execution:程序执行 Reset button:复原按钮 positive temperature coefficient:正温度系数 variable resistors:可变电阻器 cermet resistor:金属陶瓷电阻器output resistance:输出电阻distortion:失真 single amplifier:单级放大器 voltage follower:电压输出跟随器troubleshooting:发现并修理故障control panel,:控制面板 二、让我们来看看滤波器设计的基本步骤 １、打开crack的软件后，根据滤波器的设计要求，在filter type中选择滤波器的类型（Gaussian：高斯滤波器、Bessel：贝塞尔滤波器、butterworth：巴特沃斯；Chebyshev1切比雪夫1；Chebyshev2切比雪夫2；Hourglass：对三角滤波器、
Filter Solutions滤波器设计教程
一、Filter Solutions滤波器设计软件中的英文注解 Lowpass notch filters :低通陷波滤波器Order: 阶 filter circuits:滤波电路 frequency response:幅频响应Passband :通频带、传输带宽repeatedly cycle：重复周期maximum signal to noise ratio：最大信噪比 gain constants：增益系数，放大常数circuit topologies：电路拓扑结构 gain shortfall：增益不足 maximum output：最大输出功率 last stage：末级 preceding stage:前级 stage filter：分级过滤器 Gain Stage：增益级 voltage amplitude：电压振幅Component values: 元件值maximum valued: 最大值 minimum valued: 最小值 standard value:标准值 resistors: 电阻器 capacitors:电容器 operational amplifiers:运算放大器(OA) circuit board:(实验用)电路板 active filters:有源滤波器 supply currents:源电流 power supplies:电源 bypassing capacitors:旁路电容optimal:最佳的;最理想的 Gain Bandwidth:带宽增益 passive component:无源元件 active component: 有源元件 overall spread:全局;总范围Component characteristics:组件特性Modification:修改;更改 data book:数据手册 typical values:标准值;典型值 default values:省略补充 program execution:程序执行 Reset button:复原按钮 positive temperature coefficient:正温度系数 variable resistors:可变电阻器 cermet resistor:金属陶瓷电阻器output resistance:输出电阻distortion:失真 single amplifier:单级放大器 voltage follower:电压输出跟随器troubleshooting:发现并修理故障control panel,:控制面板 二、Filter Solutions滤波器设计的基本步骤 １、打开crack的软件后，根据滤波器的设计要求，在filter type中选择滤波器的类型（Gaussian：高斯滤波器、Bessel：贝塞尔滤波器、butterworth：巴特沃斯；Chebyshev1切比雪夫1；Chebyshev2切比雪夫2；Hourglass：对三角滤波器、Elliptic：椭圆滤波器、Custom：自定义滤波器、Raised Cos：升余弦滤波器、Matche：匹配滤波器、Delay：延迟滤波器）； ２、在filter class中选择滤波器的种类（低通、高通、带通、带阻）； ３、在filter Attributes中设置滤波器的阶数（Order）、通频带频率（Passband frequency）； ４、在Implementation中选择有源滤波器（active ）、无源滤波器（passive）
Q2Solutions
CIS 4301 Fall 2006 Quiz #2 Solutions Grading: Problem 1, Problem 2(1, 2) - Ravi. Problem 2(3, 4), Problem 3 – Chunglae. (40 points) Problem 1. Consider the following database schema: STUDENTS (UFID, LEVEL, SEX) EXAMS (UFID, EXAM_NUMBER, PNTS_SCORED, PNTS_POSS) PROBLEMS (UFID, EXAM_NUMBER, PROB_NUMBER, PNTS_SCORED, PNTS_POSS) The STUDENTS table lists all of the students in our class; LEVEL is either ‘Fresh’, ‘Soph’, ‘Jun’, ‘Sen’, and SEX is ‘m’ or ‘f’. EXAMS lists everyone’s score on each exam (NUMBER is the number of the exam), and PROBLEMS details how everyone did on each exam problem. Write SQL queries that compute answers to the following questions. No PL/SQL is allowed! You can create any views that you need. Your answer can either be in standard SQL or Oracle SQL. (1) Which female seniors scored greater than 90% on exam 2? (GRADING: RAVI ) SELECT S.UFID FROM STUDENTS S, EXAMS E WHERE S.UFID = E.UFID AND AND S.SEX = ‘F’ AND S.LEVEL = ‘Sen’ AND E.EXAM_NUMBER = 2 AND 10 * PNTS_SCORED > 9 * PNTS_POSS; (2) How many problems were there on exam 3? (GRADING: RAVI) SELECT COUNT ( DISTINCT PROB_NUMBER) FROM PROBLEMS WHERE EXAM_NUMBER=3; (3) For which (UFID, EXAM_NUMBER) combos is there an error? There is an error if the total points scored on the exam according to the PROBLEMS table do not match the total-points scored that is recorded in the EXAMS table. (GRADING: RAVI) CREATE VIEW EXAM_UFID_SCORE AS SELECT UFID, EXAM_NUMBER, SUM(PNTS_SCORED) SCORE FROM PROBLEMS GROUP BY UFID, EXAM_NUMBER; SELECT E.UFID, E.EXAM_NUMBER FROM EXAMS E, EXAM_UFID_SCORE EU WHERE E.UFID = EU.UFID AND E.EXAM_NUMBER = EU.EXAM_NUMBER EU.SCORE; E.PNTS_SCORED !=
THRUTUBINGSOLUTIONS井下作业工具详解
THRU TUBING SOLUTIONS井下作业工具 郑- 1526 08 2 2709 Q 285151 9548 DNH 防爆扬声器 DSP-25EEXMNT 派克马达 F12-060-MF-IV-D-000-000-P 2台 派克马达 F12-060-MF-IV-D-000-000-P 1台 TTS 连接头 2.88" Dimple Coiled Tubing Connector f/ 2" CT, 70 Mpa Working Pressure, Material: AISI 4140 HT 3个 TTS 1-2周 KSB 图尔克 电源 MK1-2RP/24VDC 2个 3M 卡洛克 26x3742 27个
TTS 延长筒 UNIVERSAL HOUSING EXTENSION F\1-11\16''MOTORF \POWER SECYION EVAC 德尔格配件 4-6周总重30KG 摩菲 防爆信息表 PV101-A-HAZ 1个 沃尔沃 3-4周 AMETEK 4-6周 Eppensteiner 油污报警器 F4.2GW0200M 阿美特克、ROCHESTER INSTRUMENT SYSTEMS 信号转换模块SC4326-NR-FC 2个信号转换器SC7405-N-R-NC 10个8-10周总重10KG DNH 扬声器 BK-560(T),3W 罗斯蒙特 高级测量变送器 3051SAM2PT2A2B11A1AR10I3M5 / 1199WDA95AFFW71DA00 + 3051SAM2ST2A2B11A2AI3
/1199WDA95AFFW71DA00 SUN 水散热放大板 790-2D24V VDO Contrinex 接近开关 DW-AS-603-M30-002 FIRETROL 充电器 LL-1580 6-8周总重5KG WTW PH计电极SenTix 41 1个 4周含税运 卡洛克 油封 21158-1678 4个 6-8周 GEA WESTFALIA 配件 摩菲 振动开关 VS94-H24-T24 1个 UFM 流量监测器 Part No.CP8 2个 4周总重14KG 其他费用40美元 FILTER-FAB/GULFGATE AI-TEK 探头 RH1512-013
雅思顾家北写作语法5lecture 1_solutions
1.很多国家都存在环境问题。 拓展：很多国家，特别是发展中国家，都存在环境问题。 正确的答案：Many countries, especially developing countries, have environmental problems. 结构：主+及物动词+宾语 正确的答案：Environmental problems exist in many countries, especially (in) developing countries. 结构：主+不及物动词；“in many countries”介宾短语充当状语；developing countries是同位语，一般紧跟在修饰的名词的后面 注： environmental problems=environmental issues 注：也可以说there exist some environmental problems in some countries. 这句话是倒装，还原成 some environmental problems exist here in some countries. 学生的答案：Many countries, especially developing countries, exist environmental problem. 错误1：exist不及物动词，不能加宾语。 错误2：problem可数名词，要加复数，或者冠词。 2.环境问题已经变的越来越严重。 拓展：在发展中国家，环境问题已经变的越来越严重。 正确的答案：Environmental problems have become increasingly serious in developing countries. 结构：主+系动词+表语“in developing countries”介宾短语充当状语； 注：increasingly 比more and more 更加好，在这个情况下 注：environmental problems=environmental issues=environmental woes Serious = severe 学生的答案：Environmental problems have been become seriously in the developing countries. 错误1：been become 两个系动词 错误2：seriously副词，不能充当表语 错误3：developing countries已经是复数，除非特指，否则不需要加定冠词the 注：使用have been becoming现在完成进行时态不恰当，因为要用一般完成时态强调结果。 3.高的失业率仍然是个问题。
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