2010山东省C语言版深入

c 2010教程C 2010是一种编程语言，广泛应用于软件开发和计算机编程领域。

它是一种面向对象的高级语言，是C语言的延伸，拥有更强大的功能和更高的效率。

在本文中，将介绍C 2010的基本概念、语法和用法。

C 2010作为一种面向对象的编程语言，具有许多优点。

首先，它具有高效的执行速度和卓越的性能。

这使得C 2010非常适合开发底层系统和需要高度优化的应用程序。

其次，C 2010拥有丰富的标准库，包括输入输出函数、字符串处理函数、数学函数等，方便开发人员快速实现各种功能。

此外，C 2010还支持多线程编程，使得程序能够更好地利用多核处理器的能力。

接下来，我们将介绍C 2010的基本语法。

C 2010的程序由一系列语句组成，每个语句以分号结尾。

变量是C 2010程序中的基本元素，用于存储数据。

要声明一个变量，需要指定其类型和名称。

C 2010支持包括整型、浮点型、字符型等多种类型的变量。

下面是一个C 2010程序的示例：#include <stdio.h>int main(){int num1 = 5;int num2 = 10;int sum = num1 + num2;printf("The sum of %d and %d is %d\n", num1, num2, sum);return 0;}在上面的示例中，我们定义了三个整型变量num1、num2和sum，并通过“+”运算符计算它们的和。

然后，使用printf函数将计算结果输出到屏幕上。

除了基本的语法规则外，C 2010还提供了许多控制结构和语句，用于实现条件判断、循环和函数调用等。

例如，if语句用于根据条件执行不同的代码块，while和for循环语句用于重复执行一段代码。

C 2010中还提供了丰富的函数库，用于实现各种功能。

例如，stdio.h库中包含了输入输出函数，math.h库中包含了数学函数等。

总之，C 2010是一种功能强大的编程语言，广泛应用于软件开发和计算机编程领域。

根据《山东省教育厅关于做好2010年普通高等教育学分互认和专科升本科工作的通知》（鲁教高字〔2009〕24号）和《关于做好2010年普通高等教育学分互认和专升本考试录取工作的通知》（鲁教高处函（2009）50号）文件（以下简称《通知》）精神，现将山东省2010年普通高等教育专升本考试工作有关事项通知如下：一、招生类别、专业和计划招生类别分为：普通专升本（含师范和高职高专）。

按照《山东省教育厅关于下达2010年普通专升本招生计划的通知》（鲁教计字〔2009〕7号），2010年山东省普通高等教育专升本招生学校、招生专业和计划见附件1，各专业考试科目见附件2。

二、报名专科（高职）生所在学校为生源学校，承担2010年专升本考试任务的学校为主考学校。

（一）报名时间普通专升本实行统一时间报名，具体时间及方式如下。

（二）报名范围及方法我省普通本科院校和专科学校（含普通专科学校、高等职业学院和成人高等学校）中的应届普通专科（高职）毕业生（含在校期间已参加过学分互认选拔但未能进入本科段学习的应届普通专科毕业生）。

此类考生在自己的生源学校报名。

具有我省辖区户籍的普通高等教育专科（高职）往届毕业生（不含成人、自学考试、网络教育）和经我省教育招生考试院录取，目前在其他省、市、自治区高校学习的2010年普通高等教育专科（高职）应届毕业生。

此类考生统一到济南大学报名，报名方式及办法见济南大学招生信息网站公布的《2010年山东省普通高等教育专升本考试报名办法》。

（三）填报志愿报名参加普通专升本考试的学生须填写《2010年山东省普通高等教育专升本考试考生报名表》（见附件3）。

填报志愿时，如果同一专业的招生学校有2所以上，考生可填报第一志愿1个，第二志愿不超过3个（平行志愿），并在“是否服从调剂录取”栏中填写“是”或“否”，若不填写，视为不服从调剂。

允许非师范类毕业生报考师范类专业，允许师范类毕业生报考非师范类专业，考生报考专业可与毕业专业不同，但每个考生只能报考师范类或非师范类中的一个专业。

2010年度山东省高等学校省级精品课程名单
2010年度山东省高等学校省级精品课程名单
2010年度山东省高等学校省级精品课程名单
2010年度山东省高等学校省级精品课程名单
2010年度山东省高等学校省级精品课程名单
2010年度山东省高等学校省级精品课程名单
2010年度山东省高等学校省级精品课程名单
2010年度山东省高等学校省级精品课程名单
2010年度山东省高等学校省级精品课程名单
2010年度山东省高等学校省级精品课程名单
2010年度筹备建设的山东省高等学校省级精品。

Problem A Phone NumberWe know that if a phone number A is another phone number B’s prefix, B is not able to be called. For an example, A is 123 while B is 12345, after pressing 123, we call A, and not able to call B.Given N phone numbers, your task is to find whether there exits two numbers A and B that A is B’s prefix.InputThe input consists of several test cases.The first line of input in each test case contains one integer N(0<N<1001), represent the number of phone numbers.The next line contains N integers, describing the phone numbers.The last case is followed by a line containing one zero.OutputFor each test case, if there exits a phone number that cannot be called, print “NO”, otherwise print “YES” instead.Sample Input Output for the Sample Input2012 012345 212 012345 0 NO YESProblem B BalloonsBoth Saya and Kudo like balloons. One day, they heard that in the central park, there will be thousands of people fly balloons to pattern a big image.They were very interested about this event, and also curious about the image.Since there are too many balloons, it is very hard for them to compute anything they need. Can you help them?Y ou can assume that the image is an N*N matrix, while each element can be either balloons or blank.Suppose element A and element B are both balloons. They are connected if: i) They are adjacent;ii) There is a list of element C 1, C 2, … , C n , while A and C 1 are connected, C 1 and C 2 are connected …C n and B are connected.And a connected block means that every pair of elements in the block is connected, while any element in the block is not connected with any element out of the block.To Saya, element A (,)xa ya and B (,)xb yb is adjacent if ||||1xa xb ya yb -+-≤ But to Kudo, element A (,)xa ya and element B (,)xb yb is adjacent if ||1xa xb -≤ and||1ya yb -≤They want to know that there’s how many connected blocks with there own definition of adjacent?InputThe input consists of several test cases.The first line of input in each test case contains one integer N (0<N ≤100), which represents the size of the matrix.Each of the next N lines contains a string whose length is N , represents the elements of the matrix. The string only consists of 0 and 1, while 0 represents a block and 1represents balloons.The last case is followed by a line containing one zero.OutputFor each case, print the case number (1, 2 …) and the connected block’s numbers with Say a and Kudo’s definition. Y our output format should imitate the sample output. Print a blank line after each test case.Sample InputOutput for the Sample Input5 11001 00100 11111 11010Case 1: 3 210010 0Problem C ClockwiseSaya have a long necklace with N beads, and she signs the beads from 1 to N. Then she fixes them to the wall to show N-1 vectors – vector i starts from bead i and end up with bead i+1.One day, Kudo comes to Saya’s home, and she sees the beads on the wall. Kudo says it is not beautiful, and let Saya make it better.She says: “I think it will be better if it is clockwise rotation. It means that to any vector i (i<N-1), it will have the same direction with vector i+1 after clockwise rotate T degrees, while 0≤T<180.”It is hard fo r Saya to reset the beads’ places, so she can only remove some beads. To saving the beads, although she agrees with Kudo’s suggestion, she thinks counterclockwise rotation is also acceptable. A counterclockwise rotation means to any vector i (i<N-1), it will have the same direction with vector i+1 after counterclockwise rotate T degrees, while 0<T≤180.”Saya starts to compute at least how many beads she should remove to make a clockwise rotation or a counterclockwise rotation.Since the necklace is very-very long, can you help her to solve this problem?InputThe input consists of several test cases.The first line of input in each test case contains one integer N (2<N≤300), which represents the number of beads.Each of the next N lines contains two integer x and y, represents the coordinate of the beads. Y ou can assume that 0<x, y<10000.The last case is followed by a line containing one zero.OutputFor each case, print your answer with the following format:If it is clockwise rotation without removing an y beads, please print “C; otherwise if it is counterclockwise rotation without removing any beads, print “CC” instead; otherwise, suppose remove at least x beads to make a clockwise rotation and remove at least y beads to make a counterclockwise rotation. If x≤y, print “Remove x bead(s), C”, otherwise print “Remove x bead(s), CC” instead.Y our output format should imitate the sample output. Print a blank line after each test case. Sample Input Output for the Sample Input31 12 23 331 12 2 CCCRemove 1 bead(s), C1 141 12 23 3 2 2Problem D ShoppingSaya and Kudo go shopping together.Y ou can assume the street as a straight line, while the shops are some points on the line.They park their car at the leftmost shop, visit all the shops from left to right, and go back to their car.Y our task is to calculate the length of their route.InputThe input consists of several test cases.The first line of input in each test case contains one integer N (0<N<100001), represents the number of shops.The next line contains N integers, describing the situation of the shops. Y ou can assume that the situations of the shops are non-negative integer and smaller than 2^30.The last case is followed by a line containing one zero.OutputFor each test case, print the length of their shopping route.Sample Input Output for the Sample Input424 13 89 3767 30 41 14 39 42 0152 70Explanation for the first sample: They park their car at shop 13; go to shop 24, 37 and 89 and finally return to shop 13. The total length is (24-13) + (37-24) + (89-37) + (89-13) = 152Problem E EmergencyKudo’s real name is not Kudo. Her name is Kudryavka Anatolyevna Strugatskia, and Kudo is only her nickname.Now, she is facing an emergency in her hometown:Her mother is developing a new kind of spacecraft. This plan costs enormous energy but finally failed. What’s more, because of the failed project, the government doesn’t have enough resource take measure to the rising sea levels caused by global warming, lead to an island flooded by the sea.Dissatisfied with her mother’s spacecraft and the government, civil war has broken out. The foe wants to arrest the spacecraft project’s participants and the “Chief criminal” –Kudo’s mother –Doctor T’s family.At the beginning of the war, all the cities are occupied by the foe. But as time goes by, the cities recaptured one by one.To prevent from the foe’s arrest and boost morale, Kudo and some other people have to distract from a city to another. Although they can use some other means to transport, the most convenient way is using the inter-city roads. Assuming the city as a node and an inter-city road as an edge, you can treat the map as a weighted directed multigraph. An inter-city road is available if both its endpoint is recaptured.Here comes the problem.Given the traffic map, and the recaptured situation, can you tell Kudo what’s the shortest path from one city to another only passing the recaptured cities?InputThe input consists of several test cases.The first line of input in each test case contains three integers N(0<N≤300), M (0<M≤100000) and Q (0<Q≤100000), which represents the number of cities, the numbers of inter-city roads and the number of operations.Each of the next M lines contains three integer x, y and z, represents there is an inter-city road starts from x, end up with y and the length is z. Y ou can assume that 0<z≤10000.Each of the next Q lines contains the operations with the following format:a)0 x– means city x has just been recaptured.b) 1 x y– means asking the shortest path from x to y only passing the recaptured cities.The last case is followed by a line containing three zeros.OutputFor each case, print the case number (1, 2 …) first.For each operation 0, if city x is already recaptured (that is,the same 0 x operation appears again), print “City x is already recaptured.”For each operation 1, if city x or y is not recaptured yet, print “City x or y is not available.” otherwise if Kudo can go from city x to city y only passing the recaptured cities, print the shortest path’s length; otherwise print “No such path.”Y our output format should imitate the sample output. Print a blank line after each test case.Sample Input Output for the Sample Input3 3 60 1 11 2 10 2 31 02 0 00 21 02 1 2 0 0 20 0 0 Case 1:City 0 or 2 is not available. 3No such path.City 2 is already recaptured.Problem F Fairy taleIt is said that in a school’s underground, there is a huge treasure which can meet any desire of the owner.The Spy Union (SU) is very interest in this legend. After much investigation, SU finally get the answer and let the youngest spy sneak into the school. That’s why Saya is now here.Today, Saya found the entrance eventually.She enters the entrance, and found her in a fairy-tale world.“Welcome!” says a fairy, “I am Ivan. My responsibility is to protect the treasure, and give it to the one who have the ability to own it.”Then Ivan gives Saya three problems.With your help, Saya finished the first and the second problem (Problem H and I). Here comes the third. If Saya can solve this problem, the treasure belongs to her.There is a big maze protecting the treasure. Y ou can assume the maze as an N*N matrix while each element in the matrix might be N (North), S (South), W (West) or E (East). At first, Saya is at the element (1, 1) – the north-west corner, and the treasure is at (N, N) – the south-east corner.The designer have enchant to this matrix, so that the treasure might move from time to time respecting the following rules:Suppose the treasure is in an element which marked with E. The treasure might eastward move a cell after a unit time. It is similar to S, W and N.After a unit time, all the mark will change: E to W, W to S, S to N, and N to E. In another word, suppose an element which marked with E at time 0. At time 1, it might change to W, change to S at time 2, change to N at time 3, change to E at time 4, and so on.Saya doesn't know the initial status of the marks. She is affected by this rule, but she decides to do something more.Ivan gave Saya a special prop which called Riki. With Riki’s help, she can get the position of the treasure.In each unit time, Saya will do all of the following three things:Firstly, she will check the treasure’s position with Riki.Secondly, she will move follow the designer’s magic the same with the treasure.(If no element exists in the direction of movement,the movement will be cancelled.)Thirdly, Saya can either move to one direction (N, S, E, and W) a cell or stay there. Saya prefers to be closer with the treasure; if there are many ways with the same geometrical distance, Saya prefers to stay there than move, prefer E than W, W than N, and N than S. Here we should use the position checked at the first step.Y ou are given the size of the matrix and all the marks of the elements at time 0. Y our task is to simulate Saya and the treasure’s movement, and then tell Saya the result.InputThe input consists of several test cases.The first line of input in each test case contains one integer N (0<N≤30), which represents the size of the matrix.Each of the next N lines contains a string whose length is N, represents the elements of the matrix. The string only consists of N, E, W and S.The last case is followed by a line containing one zero.OutputF or each case, print the case number (1, 2 …) and the result of Saya’s explore.If Saya can get the treasure at step x (x≤100)(that means at the begainning of time x, Saya and the treasure stay in the same cell), print “Get the treasure! At step x.”.If after simulating x(x≤100) steps, you found out that Saya can't get the treasure, print “Impossible. At step x.”If you have simulated 100 steps but don’t know whether Saya can get the treasure, print “Not sure.”Y our output format should imitate the sample output. Print a blank line after each test case. Sample Input Output for the Sample Input5 EWSSE NNENN EENNE SWSEW NSNSW0 Case 1:Get the treasure! At step 12.Q&AQ: How can I know it’s impossible for Saya to get the treasure?A: Suppose at time 5, Saya at (1, 1) while the treasure at (2, 2); at time 13, Saya at (1, 1) while the treasure at (2, 2) again. Since both Saya and the treasure go to the same place and have the same direction again, it is a loop, and they will just repeats the moves forever. So at time 13, we can judge it is impossible.Problem G Greatest NumberSaya likes math, because she think math can make her cleverer.One day, Kudo invited a very simple game:Given N integers, then the players choose no more than four integers from them (can be repeated) and add them together. Finally, the one whose sum is the largest wins the game. It seems very simple, but there is one more condition: the sum shouldn’t larger than a number M.Saya is very interest in this game. She says that since the number of integers is finite, we can enumerate all the selecting and find the largest sum. Saya calls the largest sum Greatest Number (GN). After reflecting for a while, Saya declares that she found the GN and shows her answer.Kudo wants to know whe ther Saya’s answer is the best, so she comes to you for help.Can you help her to compute the GN?InputThe input consists of several test cases.The first line of input in each test case contains two integers N (0<N≤1000) and M(0<M≤1000000000), which represent the number of integers and the upper bound.Each of the next N lines contains the integers. (Not larger than 1000000000)The last case is followed by a line containing two zeros.OutputFor each case, print the case number (1, 2 …) and the GN.Y our output format should imitate the sample output. Print a blank line after each test case.Sample Input Output for the Sample Input2 10Case 1: 810020 0Problem H Hello World!We know that Ivan gives Saya three problems to solve (Problem F), and this is the first problem.“We need a programmer to help us for some projects. If you show us that you or one of your friends is able to program, you can pass the first hurdle.I will give you a problem to solve. Since this is the first hurdle, it is very simple.”We all know that the simplest program is the “Hello World!” program. This is a problem just as simple as the “Hello World!”In a large matrix, there are some elements has been marked. For every marked element, return a marked element whose row and column are larger than the showed element’s row and column respectively. If there are multiple solutions, return the element whose row is the smallest; and if there are still multiple solutions, return the element whose column is the smallest. If there is no solution, return -1 -1.Saya is not a programmer, so she comes to you for help.Can you solve this problem for her?InputThe input consists of several test cases.The first line of input in each test case contains one integer N (0<N≤1000), which represents the number of marked element.Each of the next N lines containing two integers r and c, represent the element’s row and column. Y ou can assume that 0<r, c≤300. A marked element can be repeatedly showed.The last case is followed by a line containing one zero.OutputFor each case, print the case number (1, 2 …), and for each element’s row and column, output the result. Y our output format should imitate the sample output. Print a blank line after each test case.Sample Input Output for the Sample Input31 22 3 2 30 Case 1: 2 3-1 -1-1 -1Problem I Ivan comes again!The Fairy Ivan gave Saya three problems to solve (Problem F). After Saya finished the first problem (Problem H), here comes the second.This is the enhanced version of Problem H.There is a large matrix whose row and column are less than or equal to 1000000000. And there are three operations for the matrix:1)add: Mark an element in the matrix. The element wasn’t marked before it is marked.2)remove: Delete an element’s mark. The element was marked before the element’s markis deleted.3)find: Show an element’s row and column, and return a marked element’s row andcolumn, where the marked element’s row and column are larger than the showed element’s row and column respectively. If there are multiple solutions, return the element whose row is the smallest; and if there are still multiple solutions, return the element whose column is the smallest. If there is no solution, return -1.Of course, Saya comes to you for help again.InputThe input consists of several test cases.The first line of input in each test case contains one integer N(0<N≤200000), which represents the number of operations.Each of the next N lines containing an operation, as described above.The last case is followed by a line containing one zero.OutputFor each case, print the case number (1, 2 …) first. Then, for each “find” operation, output the result. Y our output format should imitate the sample output. Print a blank line after each test case.Sample Input Output for the Sample Input4add 2 3 find 1 2 remove 2 3 find 1 20 Case 1: 2 3-1Problem J Jerry MouseKudo and Saya are good friends, and they are always together.But today, since Saya is not here (Where is Saya? Y ou can get the answer at Problem F), Kudo fells very bored. So Kudo starts to watch TV for fun.Now, she is watching the famous cartoon “Tom and Jerry”. Jerry’s home have many entrance, he always enter his home in an entrance and get out from another.Kudo suddenly thought that what will happen if there are many Jerrys?Kudo finds out a paper box and dig many holes in the side of the box. Then, she marks every hole with an integer, representing its owner. Finally, she signs the box “Kudo’s House” as shown in Figure 1.Figure 1 The gray part represents a hole, and the numbers means its owner, i.e. the two gray parts in the lower left corner means the two holes belongs to mouse 2.Kudo think it is very interesting, so she makes a lot of boxes, and sign them “Saya’s House”, “Riki’s House”, “Lin’s House” and so on.Figure 2 The jointed gray parts represent the same hole.At last, she combines them together to make a large box as shown in Figure 2.She defined mouse A and mouse B is the same mouse if and only if:a)mouse A and mouse B are in a same house and A equals to B;b)mouse A and mouse B are in different houses, but they have a hole that combinedtogether, such as mouse 3 in Kudo’s House and mouse 7 in Lin’s House;c)There exits a lists of mouse M1, M2 … Mp, while A and M1 are the same, M1 and M2are the same … Mp and B are the same.But there is a problem: Suppose mouse 3 in Kudo’s House is the same with mouse 7 in Lin’s House, while mouse 7 in Lin’s House is the same with mouse 4 in Kudo’s House. It means that mouse 3 and 4 in Kudo’s H ouse are the same!Kudo is very confused, so she comes to you for help.Given the initial N boxes, can you tell her finally each hole belongs to whom?Here, N always equals to 1, 4, 9, 16, 25, 36, 49, 64, 81 or 100. Kudo always combines the boxes as shown in Figure 3, while n N=.Figure 3 How Kudo combines the boxes.InputThe input consists of several test cases.The first line of input in each test case contains two integers N (0<N≤100) and M (0<M≤1000), which represent the number of boxes and the number of holes in each side of the box. In every side of the N boxes，Y ou can assume that there are always M holes, and the M holes are arranged with equal spacing.Each of the next N lines containing 4M integers representing the holes on the boxes. The first M integers represent the holes on the upside, from left to right; the second M integers represent the holes on the downside, from left to right; the third M integers represent the holes on the leftward, from up to down; the forth M integers represent the holes on the rightward, from up to down. Y ou can assume that the hole number is not greater than M*2.The last case is followed by a line containing two zeros.Output=) represents the For each case, print the case number (1, 2 …) and 4*n*M integers (n Nholes on the upside, downside, leftward and rightward side of the large box, using the same format as the input file.In your answer, the holes signed by the same numbers should belong to the same mouse, while the holes signed by different numbers should belong to different mouse. The number should be an integer, starting from 1. Since there are multiply solutions, please print the one whose firstnumber is the least. If there are still multiply solutions, print the one whose second number is the least, and so on.Y our output format should imitate the sample output. Print a blank line after each test case. Sample Input Output for the Sample Input4 12 2 1 1 2 2 1 1 1 1 2 2 1 1 2 20 0 Case 1:1 2 1 2 3 4 3 4。

2010C程序教案（第10章）第一篇：2010C程序教案(第10章)【教学课题】预编译处理（第10章预编译处理，1、2节）【目的要求】了解编译预处理的特点和功能，掌握编译预处理中的宏定义，理解文件包含。

【教学重点】宏的定义与使用（无参和有参），理解带参的宏和函数的区别。

【教学难点】宏的定义与使用（无参和有参），理解带参的宏和函数的区别。

【教学方法】任务驱动法，问题引导法【教学手段】讲授+多媒体演示+学生动手【作业布置】1、定义一个带参数的宏，使两个参数的值互换，并写出程序，输入两个数作为使用宏时的实参。

输出已交换的两个值。

2、分别用函数和带参的宏，从3 个数中找出最大数。

一、提出任务，引入课题例1：#include main(){ int a……} 例2：#include #define N 100 main(){ int a[N];…… }。

一般C程序的开发执行过程：编辑→ 编译→ 连接→ 执行含有预处理命令的C程序：编辑→ 预处理→ 编译→ 连接→ 执行二、分析任务，讲授新课C提供的预处理功能主要有以下三种：宏定义、文件包含、条件编译。

分别用宏定义命令、文件包含命令、条件编译命令来实现。

为了与一般C语句相区别，这些命令以符号“ #” 开头。

（一）宏替换（P172）宏：代表一个字符串的标识符。

宏名：被定义为“宏”的标识符。

宏代换(展开)：在编译预处理时，对程序中所有出现的“宏名”，用宏定义中的字符串去代换的过程。

不带参数的宏定义A、一般形式：#define 标识符替换文本例如：#define PI3.14159 main(){ float l,r;printf(“Input a number :”);scanf(“%f”,&r);l=2*PI*r printf(“l=%fn”,l);} 说明：⑴宏名一般用大写表示，以便与变量名区分。

⑵使用宏名使程序易读，易修改。

⑶只作简单的置换，不作正确性检查。

c语言基本知识点总结山东省在计算机科学领域中，C语言被广泛应用于软件开发和系统编程。

精通C语言的基本知识点对于编程人员来说至关重要。

本文旨在总结C语言的基本知识点，并以山东省为背景进行描述。

一、基本语法C语言的基本语法包括关键字、标识符、数据类型、运算符、控制结构等。

在山东省，许多软件开发公司屡屡要求娴熟精通C语言的基本语法。

1.1 关键字C语言中的关键字包括if、else、while、for、switch等，它们是预定义的特殊单词，具有特殊的含义和功能。

1.2 标识符标识符是由程序员定义的名称，用于表示变量、函数、数组和其他用户自定义的对象。

在山东省，常见的标识符定名规则是使用小写字母和下划线，具有描述性，遵循CamelCase或snake_case定名风格。

1.3 数据类型C语言中的数据类型包括整型、浮点型、字符型和指针型等。

在山东省的编程实践中，屡屡使用int、float、char和void等数据类型。

1.4 运算符C语言中的运算符包括算术运算符、比较运算符、逻辑运算符和赋值运算符等。

在山东省，开发人员屡屡使用运算符进行数学计算、逻辑操作和变量赋值。

1.5 控制结构C语言中的控制结构包括条件语句和循环语句。

条件语句如if-else用于依据条件执行不同的代码块；循环语句如while 和for用于重复执行一段代码。

在山东省，开发人员屡屡使用控制结构实现复杂的程序逻辑。

二、函数和数组函数和数组是C语言中常用的观点，它们在山东省的软件开发中发挥着重要作用。

2.1 函数函数是一段预定义的可重用代码块，用于执行特定的任务。

在C语言中，函数由函数名、参数列表、返回值类型和函数体组成。

在山东省的软件开发中，函数屡屡用于实现程序的模块化和代码的复用。

2.2 数组数组是一种数据结构，用于存储一系列相同类型的元素。

在C 语言中，数组由固定大小的连续内存空间组成。

在山东省的编程实践中，数组常用于存储和处理大量的数据。

2010 年9 月全国计算机等级考试二级笔试试卷C 语言程序设计(附答案）（考试时间90 分钟，满分100 分）一、选择题（(1）—(10）、（21）-（40）每题2 分，（11）—（20）每题1 分,共70 分) 下列各题A）、B)、C）、D）四个选项中，只有一个选项是正确的，请将正确的选项填涂在答题卡相应位置上，答在试卷上不得分。

(1）下列叙述中正确的是A)线性表的链式存储结构与顺序存储结构所需要的存储空间是相同的B）线性表的链式存储结构所需要的存储空间一般要多于顺序存储结构C)线性表的链式存储结构所需要的存储空间一般要少于顺序存储结构D）上述三种说法都不对（2）下列叙述中正确的是A)在栈中,栈中元素随栈底指针与栈顶指针的变化而动态变化B)在栈中，栈顶指针不变,栈中元素随栈底指针的变化而动态变化C）在栈中，栈底指针不变，栈中元素随栈顶指针的变化而动态变化D）上述三种说法都不对（3）软件测试的目的是A）评估软件可靠性B）发现并改正程序中的错误C）改正程序中的错误D）发现程序中的错误（4)下面描述中，不属于软件危机表现的是A）软件过程不规范B)软件开发生产率低C）软件质量难以控制C）软件成本不断提高（5）软件生命周期是指A）软件产品从提出、实现、使用维护到停止使用退役的过程B）软件从需求分析、设计、实现到测试完成的过程C）软件的开发过程D)软件的运行维护过程（6）面向对象方法中，继承是指A）一组对象所具有的相似性质B）一个对象具有另一个对象的性质C)各对象之间的共同性质D）类之间共享属性和操作的机制（7)层次型、网状型和关系型数据库划分原则是A）记录长度B）文件的大小B）联系的复杂程度D）数据之间的联系方式（8）一个工作人员可以使用多台计算机,而一台计算机可被多个人使用，则实体工作人员与实体计算机之间的联系是A)一对一B）一对多C）多对多D）多对一（9）数据库设计中反映用户对数据要求的模式是A）内模式B）概念模式C）外模式D)设计模式（10）有三个关系R、S 和T 如下：则由关系R 和S 得到关系T 的操作是A）自然连接B）交C)投影D）并(11）以下关于结构化程序设计的叙述中正确的是A）一个结构化程序必须同时由顺序、分支、循环三种结构组成B）结构化程序使用goto 语句会很便捷C)在C 语言中，程序的模块化是利用函数实现的D）由三种基本结构构成的程序只能解决小规模的问题（12）以下关于简单程序设计的步骤和顺序的说法中正确的是A)确定算法后，整理并写出文档，最后进行编码和上机调试B）首先确定数据结构，然后确定算法，再编码，并上机调试，最后整理文档C）先编码和上机调试，在编码过程中确定算法和数据结构，最后整理文档D)先写好文档，再根据文档进行编码和上机调试,最后确定算法和数据结构（13）以下叙述中错误的是A）C 程序在运行过程中所有计算都以二进制方式进行B）C 程序在运行过程中所有计算都以十进制方式进行C）所有C 程序都需要编译链接无误后才能进行D)C 程序中整型变量只能存放整数,实型变量只能存放浮点数(14）有以下定义:int a;long b；double x，y;则以下选项中正确的表达式是A）a%（int）（x—y）B)a=x!=y；C）（a*y)％b D）y=x+y=x(15）以下选项中能表示合法常量的是A)整数：1,200 B）实数：1。

2012年3月全国计算机二级C语言笔试真题及详细解析（考试时间90分钟，满分100分）一、选择题1、下列叙述中正确的是：A、循环队列是队列的一种顺序存储结构B、循环队列是队列的一种链式存储结构C、循环队列是非线性结构D、循环队列是一直逻辑结构2、下列叙述中正确的是A、栈是一种先进先出的线性表B、队列是一种后进先出的线性表C、栈和队列都是非线性结构D、以上三种说法都不对3、一棵二叉树共有25个节点，其中5个时子节点，那么度为1的节点数为A、4B、6C、10D、164、在下列模式中，能够给出数据库物理存储结构与物理存取方法的是A、内模式B、外模式C、概念模式D、逻辑模式5、在满足实体完整性约束的条件下A、一个关系中可以没有候选关键词B、一个关系中只能有一个候选关键词C、一个关系中必须有多个候选关键词D、一个关系中应该有一个或者多个候选关键词6、有三个关系R、S和T如下：则由关系R和S得到关系T的操作是A、自然连接B、并C、差D、交7、软件生命周期中的活动不包括A、软件维护B、市场调研C、软件测试D、需求分析8、下面不属于需求分析阶段任务的是A、确定软件系统的功能需求B、确定软件系统的系统的系能需求B、制定软件集成测试计划D、需求规格说明书审评9、在黑盒测试方式中，设计测试用例的主要根据是A、程序外部功能B、程序内部逻辑C、程序数据结构D、程序流程图10、在软件设计中不使用的工具是A、系统结构图B、程序流程图C、PAD图D、数据流图（DFD图）11、针对简单程序设计，以下叙述的实施步骤正确的是A、确定算法和数据结构、编码、调试、整理文档B、编码、确定算法和数据结构、调试、整理文档C、整理文档、确定算法和数据结构、编码、调试D、确定算法和数据结构、调试、编码、整理文档12、关于C语言中数的表示，以下叙述正确的是A、只有整型数在允许范围内能精确无误的表示，实型数会有误差B、只要在在允许范围内整型和实型都能精确表示C、只有实型数在允许范围内能精确无误的表示，整型数会有误差D、只有八进制表示的数在不会有误差13、以下关于算法叙述错误的是A、算法可以用伪代码、流程图等多种形式来描述B、一个正确的算法必须有输入C、一个正确的算法必须有输出D、用流程图可以描述的算法可以用任何一种计算机高级语言编写成程序代码14、以下叙述错误的是A、一个C程序可以包含多个不同名的函数B、一个C程序只能有一个主函数C、C程序在书写时，有严格的缩进要求，否则不能编译通过D、C程序的主函数必须用main作为函数名15、设有以下语句Char ch1,ch2, scanf（”%c%c”,&ch1,&ch2）;若要为变量ch1和ch2分别输入字符A和B，正确的输入形式应该是A、A和B之间用逗号间隔B、A和B之间不能有任何间隔符C、A和B之间可以用回车间隔D、A和B之间用空格间隔16、以下选项中非法的字符常量是A、’\102’B、’\65’C、’\xff’D、’\019’17、有以下程序#include <sthio.h>Main(){Int A=0,B=0,C=0;C=(A-=A-5);(A=B,B+=4);Printf(“%d, %d, %d\n”,A,B,C)}程序运行后输出的结果是A 0,4,5B 4,4,5C 4,4,4D 0,0,018、设变量均已正确定义并且赋值，以下与其他三组输出结构不同的一组语句是A、x++; printf((“%d\n”,x);B、n=++x; printf((“%d\n”,n);C、++x; printf((“%d\n”,x);D、n=x++; printf((“%d\n”,n);19、以下选项中，能表示逻辑值“假”的是A 1B 0.000001C 0D 100.020、有以下程序#include <sthio.h>Main(){ int a；Scanf(“%d”,&a);If(a++<9) printf((“%d\n”,a);Else printf((“%d\n”,a--);}程序运行时键盘输入9<回车>，则输出的结构是A、10 B 11 C 9 D 821、有以下程序#include <sthio.h>Main(){int s=0,n;For (n=0;n<3;n<++){switch(s){ case 0;Case 1;s+=1;Case 2;s+=2;break;Case 3;s+3;Case 4;s+=4;}printf((“%d\n”,s);}}程序运行后的结果是A 1,2,4B 1,3,6C 3,10,14D 3,6,1022、若k是int类型变量，且有以下for语句For（k=-1；k<0;k++）printf(****\n”);下面关于语句执行情况的叙述中正确的是A、循环体执行一次B、循环体执行两次C、循环体一次也不执行D、构成无限循环23、有以下程序#include <sthio.h>Main(){char A,B,C;B=’1’;C=’A’For(A=0;A<6;A++){if(A%2) putchar（B+A）;Else putchar（C+A）；}}程序运行后输出的结果是A 1B3D5FB ABCDFEC A2C4E6D 12345624、设有如下定义语句Int m[ ]={2,4,6,8},*k=m;以下选项中，表达式的值为6的是A *(k+2)B k+2C *k+2D *k+=225、fun函数的功能是：通过键盘输入给x所指的整型数组所有元素赋值。

计算机程序设计（C）教案2010年~2011年第二学期计算机学院〃基础教研室教材：《大学C/C++语言程序设计》主编：阳小华、马淑萍出版社：电子工业出版社《大学C/C++语言程序设计实验教程》主编：阳小华、邹腊梅出版社：电子工业出版社参考书：1.《C程序设计》（第二版）谭浩强著，清华大学出版社2.《湖南省普通高等学校计算机水平等级考试辅导教材（第二版）》邹北骥等编著3. 其他相关书籍。

第 1 讲课程名称：《计算机程序设计（C）》授课章节：第1章程序设计概述授课对象：2010级授课学时：2授课方式：课堂讲授一、教学目的与要求：使学生熟悉和掌握C语言的主要特征，C语言的程序结构和C语言的运行环境。

二、讲授内容：C语言的主要特征，C语言的程序结构和C语言的运行环境、算法、。

三、教学重点：C语言的程序结构、算法、。

四、教学难点：C语言的程序结构和C语言的运行环境。

五、教具：多媒体教学。

六、课后作业：教材习题1七、预习内容：教材P.1 ~ 10课程名称：《计算机程序设计（C）》授课章节：第2章 C语言基础2.1 基本概念2.2 C语言的数据类型授课对象：2010级授课学时：2授课方式：课堂讲授一、教学目的与要求：了解和掌握C语言基本概念及数据类型。

二、讲授内容：C语言基本概念、字符集、标识符、关键字、常量、变量以及C语言的数据类型。

三、教学重点：C语言基本概念。

四、教学难点：C语言的数据类型。

五、教具：多媒体教学。

六、课后作业：教材习题2。

七、预习内容：教材P.12 ~ 23课程名称：《计算机程序设计（C）》授课章节：第2章 C语言基础2.3 运算符与表达式2.4 小结授课对象：2010级授课学时：2授课方式：课堂讲授一、教学目的与要求：了解和掌握C语言运算符与表达式二、讲授内容：C语言运算符与表达式。

三、教学重点：C语言表达式。

四、教学难点：C语言运算优先级。

五、教具：多媒体教学。

六、课后作业：教材习题2。

《Visual C++ 2010入门教程》写在前面在我还在上学的时候，我选择了C++，最初我用VC6作为我的IDE，我看过很多本C++的教材，有的适合我，有的不适合我，其中有一本叫《Visual C++ 2005入门经典》的书帮了我不少的忙。

因为通常的C++教材都只会介绍C++的语法什么的，很少会告诉我们如何去编译、运行，告诉我们什么是控制台程序，什么事Win程序，什么是GUI程序，C++能干什么，VC和C++的区别是什么。

现在有很多的朋友应该也有这些问题吧？学C++用C++也有几年了，算不上熟悉，算是初窥门径吧，我想我应该做点什么帮助一下那些和曾经的我一样困惑的朋友，特别是学生朋友，告诉他们他们所困惑的问题的答案。

记得我学C++的时候，没有人教，有的时候也走了不少弯路，甚至连调试也不会，也不知道可以通过看调用堆栈看调用次序，还自己慢慢的去搜索，好傻啊。

接下来我会做一个《Visual C++ 2010入门教程》系列，用来帮助初学者。

刚开始学的时候是很痛苦的，这个我深有体会，特别是身边还没有人能够指导一二的。

内容主要涵盖在Windows下面使用C++进行开发的常见内容，Visual Studio 2010的使用，如何创建新项目，如何调试，如果配置项目属性等等，另外还会介绍Visual C++ 2010中新加如的一些内容，包括一些新的STL组建，一些新的语法支持等等。

由于本人水平有限，其中难免有错误，希望大家谅解，如果大家有发现问题还请务必及时指出来，否则误导了他人我就罪不容恕了。

注意，本教程非C++教程，不会教你C++，只会教你如何使用 Visual C++ 2010去练习去学习其它C++教材上面的程序。

因此建议大家认真的去看其他的C++教程，在使用Visual C++ 2010实践的时候如果遇到问题可以到这里来参考。

推荐《C++ Primer》、《C++程序设计语言》《Visual C++ 2008入门经典》。