参观2005中国国际通信展览会感想

随着通信技术日益进步,一个可以随时随地,以各种方式自由沟通,享受信息技术带来的无限自由的生活时代已经到来。
通过这次参观2005中国国际通信展览会,我深切的体会到了通信技术的魅力,特别是这次通信展上各大厂商展出3G技术和产品更是吸引大量眼球。

3G技术的研发和商用是当前国际通信行业的热点,在该领域具有领先地位的国内外知名企业纷纷参展。日本NTT-DoCoMo是日本最大的移动通信运营商,在全球最早实现WCDMA商用,此次展出了3G商业运营的新模式;上海贝尔阿尔卡特以“创新宽带生活”、“用户核心宽带”的主题,在展会上推出了支持全标准的3G解决方案,从成熟的3G、规划的3G、业务的3G和服务的3G四个方面全面介绍解决方案;大唐集团推出具有中国知识产权的名扬海外的3G核心技术——TD-SCDMA;朗讯科技以“真正的价值源于真正的融合”为主题,全面展示领先的融合网络理念和解决方案,包括时尚融合业务、3G网络及应用等。英华达公司作为3G“TD-SCDMA产业联盟”的新成员首次亮相上海展会。

上海移动展出的一辆轿车与一般的车辆看似无多少差别,但是介绍人员展示了无线功能,让人刮目相看。将手机放在车内,将出发地和目的地输入后,即会出现语音提示,“前面左转……”不认路的驾驶员也能轻松上路。具介绍这一系统还能提供道路信息,避开拥堵。如果车辆受袭,被打开车门或击损,会立即报警,可设5个电话依次报告。3G网络高带宽给手机电视带来高质量的视频内容,相当于带了一台可移动电视机,一路上可以看体育节目、新闻节目、娱乐节目,访问媒体门户网站。

上海移动还展示了移动服务“电子回执”让人很称奇。客户在网上订票后,并以信用卡支付后,可通过移动数据通道,获得票务信息,包括场次、座位号,或航班、日期等。当客户将手机显示的票务信息在终端识读机一照,所有信息还原,可以完成传统意义的验票。此外还有手机钱包,通过移动接入,提供账户的金融信息服务,可以理财、转账、查询、缴费等。

上海联通也把准3G的CDMA1X带来展会现场,CDMA1X可以提供高速无线互联网接入和丰富的数据业务。据介绍联通可以在极短的时间内,花较少成本,将CDMA1X的网络直接升级到3G,可支撑联通未来的网上电影、无线电视、在线音乐、移动邮箱等业务。联通已得到总部的批准,将设立国内首个商用型3G试验网。这张商用试验网计划在今年七八月形成公众目标,年底实现区域覆盖。

今年在上海举行的世乒赛上,联通的“准”3G网络已经小试身手,该网络从虹桥机场、漕河泾到上海体育场,覆盖全市6个区域。用户通过CDMA20001X增强型手机,能成功地收看CCTV1、东方卫视、上视体育频道的实况节目转播。

日本NTT在展会上展示了基于FOMA(第三代网络)和I-Mode(移动互联网平台)的多种业务应用。并介绍了已经有超过1100万用户使用FOMA业务,并且这个数字还在不断增长,FOMA提供用于移动多媒体服务的高速分组交换模式和用于视频会议的电路交换模式,此项服务已经覆盖整个日本。

第一代手机为模拟制式,第二代手机为GSM、TDMA等数字手机,而所谓第三代手机,则是泛称能够将语音通信和多媒体通信相结合的新一代移动通信系统,其可能的增值服务将包括图像、音乐、网页浏览、电话会议以及其它一些信息服务。

第三代手机的名称繁多,国际电联称之为“IMT-2000”,欧洲的电信业巨头们则称其为“UMTS”通用移动通信系统,第三代手机可能应用的技术标准WCDMA、CDMA2000、TD-SCDMA等也在一些场合被作为第三代手机的代称,而更笼统地称呼则为“3G”:The Third Generation。

3G手机完全是通信工业和计算机工业相融合的产物,和此前的手机相比差别实在是太大了,因此越来越多的人开始称呼这类新的移动通信产品为“个人通信终端”。即使是对通信业最外行的人也可从外形上轻易地判断出一台手机是否是“第三代”:第三代手机都有一个超大的彩色显示屏,往往还是触摸式的!

许多人和第三代手机的第一次亲密接触始于广告。诺基亚曾在电视上播放过未来手机的概念样机,一个快乐的旅游者手持3G手机徘徊在陌生的城市,手机里显示的街道地图指引他顺利抵达目的地。

3G手机能干什么?“它能给我们真正的沟通自由!”;“3G到底有多Fun”:不必经过痛苦的学习就能用3G终端和亲人、朋友、陌生人甚至与设备间共享数据、图像、视频和思想。

“块头”有点大,样子有点“土”,这就是我们在展览上看到的几部3G手机。可是别看它其貌不扬,由于3G网络传输速度极快,能传输大容量的视频和声音文件,3G手机几乎能集合互动电视、上网电脑、远程监控等许多功能,变得“越来越不像手机”,人们已经无法再将它和移动电话机直接对应起来。

然而,3G手机各种功能的实现,最终要依赖于运营商3G网络的开通。由于目前我国尚未向运营商发放3G运营牌照,3G网络建设还没有启动。但在联通展区,我们也体验一下上海联通的“准”3G网络,网络速度几乎可以与ADSL有线宽带相媲美。手机用户借助它能进行快速上网、高速收看实时电视节目,VOD点播、视频即时通信等应用。

这就是美好的3G生活,眼镜、手表、化妆盒、旅游鞋,以方便和个性为前提,任何一件你能看到的物品都有可能成为3G终端。当带宽和技术具备时,这个集成许多功能的终端使我们不仅可以随时随地通信,更可以双向下载传递资料、图画、影像,当然更可以和从未谋面的陌生人网上联线对打游戏。

3G时代的手机除了能高质量的完成目前手机所做的语音通信外,还能进行多媒体通信。用户可以在3G手机的触摸显示屏上直接写字、绘图,并将其传送给另一台手机,而所需时间可能不到一秒。当然,也可以将这些信息传送给一台电脑,或从电脑中下载某些信息;用户可以用3G手机直接上网,查看电子邮件或浏览网页;不少型号的3G手机自带摄像头,这将使用户可以利用手机进行电脑会议,甚至使数字相机成为一种“多余”。

具备强大功能的基础是3G手机极高的数据传输速度,目前的GSM移动通信网的传输速度为每秒9.6K字节,而第三代手机最终可能达到的数据传输速度将高达每秒2M字节。而为此做支撑的则是互联网技术充分糅合到3G手机系统中,其中最重要的就是数据打包技术。在现有GSM上应用数据打包技术发展出的GPRS目前已可达到每秒384K字节的传输速度,这相当于D-ISDN传输速度的两倍。3G手机支持高质量的话音,分组数据,多媒体业务和多用户速率通讯,将大大扩展手机通讯的内涵。

全球3G网络的技术标准有我国自主开发的TD-SCDMA、美国的CDMA2000和欧洲的WCDMA技术标准。我国第三代移动电话3G标准将由市场决定,而不是政府。我国TD-SCDMA技术标准已经得到包括摩托罗拉在内的多家公司的认可。TD-SCDMA技术比WCDMA、CDMA2000有优势,它能充分利用空气频谱传输信息,因此传递速度特快,成本也较便宜,下载互联网信息也较快。

通过这些介绍,我们不仅对3G的技术知识、3G的商业运用、通信专业有了大致的了解,也是使我们对中国3G和通信专业的发展方向有了很大的信心。

中午无聊!!!无聊!!!

上午一大早起来载了《七剑》,刚看完,感觉也不怎么样,现在很少碰到有感觉的片子了,看来我还是喜欢比较感人的片子,听歌也只听写比较抒情的、Blues的,品味如就。说到歌,蔡淳佳又发新专辑了,听了听,感觉不错,从第一张专辑到现在风格还是一样。
  无聊,开了MSN(一般只开Popo用MSN插件,看不到Space的更新)看了看朋友的Space,看来大家暑期的活动还是挺丰富的。我还是比较懒,同学的聚会由于天气炎热没去,看看外面,火火的太阳,马上又要红色预警信号了吧,哎~~~这鬼天,怎么不像军训时候一样也来个什么什么的台风,刮一刮,爽一下。
  昨天参加了TopCoder(R) Single Round Match 258 做了Point 250 和Point 1000 的题目,Point 500 的看了半天没看懂,大概讲的是AutoLoan 奥的公司汽车贷款利息计算方面的问题,时间也不够,如果是像高中时候的都是中文的应该没什么问题的。MD,Point 250 的到是过了System Test,主要是Point 1000 的题,竟然给别人给Challenge Successfully了,差点气的吐血,看来还是得好好努力啊,期待16号中午开始的TCO Algorithm Competition 希望有好的成绩,对了要查查时间东部时间Noon是什么时候,北京时间好像是半夜三更哦,我晕。。。
  查到了,死了,是17号0:00 。。。不管了,努力,努力!!!

背景音乐:马郁 下辈子如果我还记得你

TopCoder(R) Single Round Match 258

Problem Statement for ClassScores

Problem Statement

    

A teacher has just finished grading the test papers for his class. To get an idea of how difficult the test was, he would now like to determine the most common score on the test. In statistics, this is called the "mode" of a set of data points. For instance, if the scores were {65, 70, 88, 70}, then the mode would be 70, since it appears twice while all others appear once.

Sometimes, in the case of a tie, the mode will be more than one number. For instance, if the scores were {88, 70, 65, 70, 88}, then the mode would be {70, 88}, since they both appear most frequently.

You are given a int[] scores. You are to return a int[] representing the mode of the set of scores. In the case of more than one number, they should be returned in increasing order.

 

Definition

    
Class: ClassScores
Method: findMode
Parameters: int[]
Returns: int[]
Method signature: int[] findMode(int[] scores)
(be sure your method is public)
    
 
 

Constraints

scores will contain between 1 and 50 elements, inclusive.
Each element of scores will be between 0 and 100, inclusive.
 

Examples

0)  
    
The first example from the problem statement.
1)  
    
The second example from the problem statement.
2)  
    
With no duplicates, all of the elements are the most frequent (appearing once each).

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2005, TopCoder, Inc. All rights reserved.

 


 

 

public class ClassScores {

public int[] findMode(int[] scores) {
  
int[] count = new int[101];
  
for (int i = 0; i < scores.length; i++)
    count[scores[i]]
++;
  
for (int i = scores.length; i ≥ 1; i{
    
int c = 0;
    
for (int j = 0; j ≤ 100; j++)
      
if (count[j] == i)
        c
++;
    
if (c > 0{
      
int p = 0;
      
int[] ret = new int[c];
      
for (int j = 0; j ≤ 100; j++)
        
if (count[j] == i) {
          ret[p] 
= j;
          p
++;
        }

      
return ret;
    }

  }

  
return new int[0];
}


}


 


 

 

Problem Statement for AutoLoan

Problem Statement

    

Auto dealerships frequently advertise tempting loan offers in order to make it easier for people to afford the "car of their dreams". A typical sales tactic is to show you various cars, and then talk in terms of what your monthly payment would be, to say nothing of how much you are actually paying for the car, how much interest you pay, or how long you have to make payments.

A typical auto loan is calculated using a fixed interest rate, and is set up so that you make the same monthly payment for a set period of time in order to fully pay off the balance. The balance of your loan starts out as the sticker price of the car. Each month, the monthly interest is added to your balance, and the amount of your payment is subtracted from your balance. (The payment is subtracted after the interest is added.) The monthly interest rate is 1/12 of the yearly interest rate. Thus, if your annual percentage rate is 12%, then 1% of the remaining balance would be charged as interest each month.

You have been checking out some of the cars at your local dealership, TopAuto. An excited salesman has just approached you, shouting about how you can have the car you are looking at for a payment of only monthlyPayment for only loanTerm months! You are to return a double indicating the annual percentage rate of the loan, assuming that the initial balance of the loan is price.

 

Definition

    
Class: AutoLoan
Method: interestRate
Parameters: double, double, int
Returns: double
Method signature: double interestRate(double price, double monthlyPayment, int loanTerm)
(be sure your method is public)
    
 
 

Notes

Because of the way interest is compounded monthly, the actual interest accrued over the course of a year is not necessarily the same as (balance * yearly interest rate). In fact, it's usually more.
In a real situation, information like this would typically need to be disclosed, but since you aren't at a point of signing any paperwork, the salesman has no legal obligation to tell you anything.
The return value must be within 1e-9 absolute or relative error of the actual result.
 

Constraints

price will be between 1 and 1000000, inclusive.
monthlyPayment will be between 0 and price / 2, inclusive.
loanTerm will be between 1 and 600, inclusive.
The resulting interest rate will be between 0 and 100, inclusive.
 

Examples

0)  
    
Noting that 68 payments of 100 equals the total price of 6800, so there is no interest.
1)  
    
Here, we do pay a little interest. At 9.562% annual interest, that means each month we pay 0.7968% of the balance in interest. Our payment schedule looks like this:
2)  
    
This is similar to what purchasing a new car with no money down might look like, if you make payments for 4 years.

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2005, TopCoder, Inc. All rights reserved.

 


 

 

public class AutoLoan {

private double amort(double principal, double payment, int term, double interest) {
  
double m = interest / 1200;
  
if (principal * m > payment)
    
return 1;
  
for (int i = 0; i < term; i++)
    principal 
= principal * (1 + m)  payment;
  
return principal;
}


public double interestRate(double price, double monthlyPayment, int loanTerm) {
double ret = 0;
double inc = 1000000000;
while (inc ≥ 1.0E-18{
  
double d = amort(price, monthlyPayment, loanTerm, ret + inc);
  
if (d ≤ 0{
    ret 
+= inc;
  }

  inc 
/= 2.0;
}


return ret;

}


}


 


Problem Statement for MissileTarget

Problem Statement

    

You are working for a defense agency that is testing the accuracy of a new missile guidance system. As part of this effort, several missiles have been fired off. Each missile fired was programmed with the same target coordinates, although the actual points of impact vary.

Your task is to determine the "best fit" point to describe the location where the missiles actually landed. To determine how well a point describes the location, calculate the cartesian distance from the point to each of the landing points. Then, total the sum of the squares of these distances. The best fit point is the point that minimizes this sum.

You are given int[]s x and y, both containing the same number of elements, where the i-th element of x and the i-th element of y describe the coordinates of the i-th missile landing point. You are to return a int[] with exactly two elements, describing the coordinates of the lattice point (point with integral coordinates) that is closest to the "best fit" point. The first element should be the x-coordinate, and the second element should be the y-coordinate.

 

Definition

    
Class: MissileTarget
Method: bestFit
Parameters: int[], int[]
Returns: int[]
Method signature: int[] bestFit(int[] x, int[] y)
(be sure your method is public)
    
 
 

Notes

The cartesian distance between two points (x1, y1) and (x2, y2) is defined as Sqrt((x2-x1)^2 + (y2-y1)^2).
The return value must be within 1e-9 absolute or relative error of the actual result.
 

Constraints

x will contain between 1 and 50 elements, inclusive.
x and y will contain the same number of elements.
Each element of x will be between -1000000 and 1000000, inclusive.
Each element of y will be between -1000000 and 1000000, inclusive.
The actual (possibly non-lattice) best fit point will be at least 1e-2 closer to the correct return value than to any other lattice point.
 

Examples

0)  
    
These three impacts are all pretty close to the origin, and sure enough, the origin is the best fit point.
1)  
    
With only one point, it is its own best fit.
2)  
    
With only two points, the best fit is the midpoint between the two.
3)  
    
 
4)  
    
In this case, notice that the actual best fit point possible is (5.333, 0). If we look at lattice points only, then our best fit is (6, 0), however, we are interested in the lattice point that is closest to the actual best fit point, so we return (5, 0).

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2005, TopCoder, Inc. All rights reserved.


 

This is another problem that is fairly easily solved with a bit of grunt work to calculate out the desired values. Since we are looking for the lattice point that is closest to our best fit, our best bet is to first calculate the location of the actual best fit point (using floating point, that is), and then find the closest lattice point.

To find the best fit point, we one important observation: calculating the best fit x-coordinate and the best fit y-coordinate separately will give us our best fit point. Why? Since the scoring of a point as being best fit is based upon the sum of the squares of the distances from each of the points, we see that:

score = sum(d^2) = sum(sqrt((xx0)^2 – (yy0)^2)^2)
  = sum((xx0)^2 + (yy0)^2)
  = sum((xx0)^2) + sum((yy0)^2)

So, to minimize the score, it suffices to minimize each sum separately.

To minimize each sum, a ternary search works well. However, in this case, if you were inclined to do the mathematical gruntwork, then you found a nice shortcut. The average of the x-coordinates will give you the x-coordinate of the best fit point, and the same goes for the y-coordinates. (Why? Hint: Use calculus to prove where the minimum value is.)

Either way, once you have the location of the best fit point it's just simply a matter of finding the closest lattice point, and the easiest way to do this is by rounding. (Note the constraints were intended to prohibit the case where a point was equidistant from multiple lattice points.)