微软2014实习生招聘及技术类秋令营招聘onlinetest编程题

第一篇：微软2014实习生招聘及技术类秋令营招聘onlinetest编程题

1.String reorder Time Limit: 10000ms Case Time Limit: 1000ms Memory Limit: 256MB

Description

For this question, your program is required to process an input string containing only ASCII characters between ‘0’ and ‘9’, or between ‘a’ and ‘z’(including ‘0’, ‘9’, ‘a’, ‘z’).Your program should reorder and split all input string characters into multiple segments, and output all segments as one concatenated string.The following requirements should also be met, 1.Characters in each segment should be in strictly increasing order.For ordering, ‘9’ is larger than ‘0’, ‘a’ is larger than ‘9’, and ‘z’ is larger than ‘a’(basically following ASCII character order).2.Characters in the second segment must be the same as or a subset of the first segment;and every following segment must be the same as or a subset of its previous segment.Your program should output string “” when the input contains any invalid characters(i.e., outside the '0'-'9' and 'a'-'z' range).Input

Input consists of multiple cases, one case per line.Each case is one string consisting of ASCII characters.Output

For each case, print exactly one line with the reordered string based on the criteria above.Sample In

aabbccdd 007799aabbccddeeff113355zz 1234.89898 abcdefabcdefabcdefaaaaaaaaaaaaaabbbbbbbddddddee

Sample Out

abcdabcd 013579abcdefz013579abcdefz Abcdefabcdefabcdefabdeabdeabdabdabdabdabaaaaaaa

2.K-th string Time Limit: 10000ms Case Time Limit: 1000ms Memory Limit: 256MB

Description

Consider a string set that each of them consists of {0, 1} only.All strings in the set have the same number of 0s and 1s.Write a program to find and output the K-th string according to the dictionary order.If such a string doesn’t exist, or the input is not valid, please output “Impossible”.For example, if we have two ‘0’s and two ‘1’s, we will have a set with 6 different strings, {0011, 0101, 0110, 1001, 1010, 1100}, and the 4th string is 1001.Input

The first line of the input file contains a single integer t(1 ≤ t ≤ 10000), the number of test cases, followed by the input data for each test case.Each test case is 3 integers separated by blank space: N, M(2 <= N + M <= 33 and N , M >= 0), K(1 <= K <= 1000000000).N stands for the number of ‘0’s, M stands for the number of ‘1’s, and K stands for the K-th of string in the set that needs to be printed as output.Output

For each case, print exactly one line.If the string exists, please print it, otherwise print “Impossible”.Sample In 3 2 2 2 2 2 7 4 7 47

Sample Out

0101 Impossible 01010111011

3.Reduce inversion count Time Limit: 10000ms Case Time Limit: 1000ms Memory Limit: 256MB

Description

Find a pair in an integer array that swapping them would maximally decrease the inversion count of the array.If such a pair exists, return the new inversion count;otherwise returns the original inversion count.Definition of Inversion: Let(A[0], A[1]...A[n])be a sequence of n numbers.If i < j and A[i] > A[j], then the pair(i, j)is called inversion of A.Example: Count(Inversion({3, 1, 2}))= Count({3, 1}, {3, 2})= 2 InversionCountOfSwap({3, 1, 2})=> {

InversionCount({1, 3, 2})= 1 <--swapping 1 with 3, decreases inversion count by 1

InversionCount({2, 1, 3})= 1 <--swapping 2 with 3, decreases inversion count by 1

InversionCount({3, 2, 1})= 3 <--swapping 1 with 2 , increases inversion count by 1 }

Input

Input consists of multiple cases, one case per line.Each case consists of a sequence of integers separated by comma.Output

For each case, print exactly one line with the new inversion count or the original inversion count if it cannot be reduced.Sample In

3,1,2 1,2,3,4,5

Sample Out 1 0

4.Most Frequent Logs

Time Limit: 10000ms Case Time Limit: 3000ms Memory Limit: 256MB

Description

In a running system, there're many logs produced within a short period of time, we'd like to know the count of the most frequent logs.Logs are produced by a few non-empty format strings, the number of logs is N(1=N=20000), the maximum length of each log is 256.Here we consider a log same with another when their edit distance(see note)is = 5.Also we have a)logs are all the same with each other produced by a certain format string b)format strings have edit distance 5 of each other.Your program will be dealing with lots of logs, so please try to keep the time cost close to O(nl), where n is the number of logs, and l is the average log length.Note edit distance is the minimum number of operations(insertdeletereplace a character)required to transform one string into the other, please refer to httpen.wikipedia.orgwikiEdit_distance for more details.Input

Multiple lines of non-empty strings.Output

The count of the most frequent logs.Sample In

Logging started for id:1 Module ABC has completed its job Module XYZ has completed its job Logging started for id:10 Module ? has completed its job

Sample Out 3

第二篇：微软2014实习生招聘及技术类秋令营招聘onlinetest编程题

1.String reorder

Time Limit: 10000ms

Case Time Limit: 1000ms

Memory Limit: 256MB

Description

For this question, your program is required to process an input string containing only ASCII characters between ‘0’ and ‘9’, or between ‘a’ and ‘z’(including ‘0’, ‘9’, ‘a’, ‘z’).Your program should reorder and split all input string characters into multiple segments, and output all segments as one concatenated string.The following requirements should also be met,1.Characters in each segment should be in strictly increasing order.For ordering, ‘9’ is larger than ‘0’, ‘a’ is larger than ‘9’, and ‘z’ is larger than ‘a’(basically following ASCII character order).2.Characters in the second segment must be the same as or a subset of the first segment;and every following segment must be the same as or a subset of its previous segment.Your program should output string “” when the input contains any invalid characters(i.e., outside the '0'-'9' and 'a'-'z' range).Input

Input consists of multiple cases, one case per line.Each case is one string consisting of ASCII characters.Output

For each case, print exactly one line with the reordered string based on the criteria above.Sample In

aabbccdd

007799aabbccddeeff113355zz

1234.89898

abcdefabcdefabcdefaaaaaaaaaaaaaabbbbbbbddddddee

Sample Out

abcdabcd

013579abcdefz013579abcdefz

Abcdefabcdefabcdefabdeabdeabdabdabdabdabaaaaaaa

2.K-th string

Time Limit: 10000ms

Case Time Limit: 1000ms

Memory Limit: 256MB

Description

Consider a string set that each of them consists of {0, 1} only.All strings in the set have the same number of 0s and 1s.Write a program to find and output the K-th string according to the dictionary order.If such a string doesn’t exist, or the input is not valid, please output “Impossible”.For example, if we have two ‘0’s and two ‘1’s, we will have a set with 6 different strings, {0011, 0101, 0110, 1001, 1010, 1100}, and the 4th string is 1001.Input

The first line of the input file contains a single integer t(1 ≤ t ≤ 10000), the number of test cases, followed by the input data for each test case.Each test case is 3 integers separated by blank space: N, M(2 <= N + M <= 33 and N , M >= 0), K(1 <= K <= 1000000000).N stands for the number of ‘0’s, M stands for the number of ‘1’s, and K stands for the K-th of string in the set that needs to be printed as output.Output

For each case, print exactly one line.If the string exists, please print it, otherwise print “Impossible”.Sample In2 22 77 47

Sample Out

0101

Impossible

01010111011

3.Reduce inversion count

Time Limit: 10000ms

Case Time Limit: 1000ms

Memory Limit: 256MB

Description

Find a pair in an integer array that swapping them would maximally decrease the inversion count of the array.If such a pair exists, return the new inversion count;otherwise returns the original inversion count.Definition of Inversion: Let(A[0], A[1]...A[n])be a sequence of n numbers.If i < j and A[i] > A[j], then the pair(i, j)is called inversion of A.Example:

Count(Inversion({3, 1, 2}))= Count({3, 1}, {3, 2})= 2

InversionCountOfSwap({3, 1, 2})=>

{

InversionCount({1, 3, 2})= 1 <--swapping 1 with 3, decreases inversion count by 1

InversionCount({2, 1, 3})= 1 <--swapping 2 with 3, decreases inversion count by 1

InversionCount({3, 2, 1})= 3 <--swapping 1 with 2 , increases inversion count by 1

}

Input

Input consists of multiple cases, one case per line.Each case consists of a sequence of integers separated by comma.Output

For each case, print exactly one line with the new inversion count or the original inversion count if it cannot be reduced.Sample In

3,1,2

1,2,3,4,5

Sample Out

4.Most Frequent Logs

Time Limit: 10000ms

Case Time Limit: 3000ms

Memory Limit: 256MB

Description

In a running system, there're many logs produced within a short period of time, we'd like to know the count of the most frequent logs.Logs are produced by a few non-empty format strings, the number of logs is N(1=N=20000), the maximum length of each log is 256.Here we consider a log same with another when their edit distance(see note)is = 5.Also we have a)logs are all the same with each other produced by a certain format string b)format strings have edit distance5 of each other.Your program will be dealing with lots of logs, so please try to keep the time cost close to O(nl), where n is the number of logs, and l is the average log length.Note edit distance is the minimum number of operations(insertdeletereplace a character)required to transform one string into the other, please refer to httpen.wikipedia.orgwikiEdit_distance for more details.Input

Multiple lines of non-empty strings.Output

The count of the most frequent logs.Sample In

Logging started for id:1

Module ABC has completed its job

Module XYZ has completed its job Logging started for id:10

Module ? has completed its job

Sample Out

第三篇：腾讯2010实习生招聘笔试题目(技术类)

1.第一题是软件工程中关于聚合在UML中的画法

空心菱靠在整体类上

2.有一个题是求一个图中最小生成树的权值总和

3.有一个是根据后序和中序求树的先序遍历

一个是关于sizeof的题目，char A[]=“hello”，求sizeof(A),这一问答案是6，再就是 char *p=A;求sizeof（p），答案是4，再就是 int a[100]求sizeof(a)答案是400

4.有一个题目是求散列表存储数据之后的数组里面各单元存储的数据内容

5.有一个是判断关于RISC（精简指令）的特点

6.有一个是选择进程的执行顺序防止死锁

7.当判断网络故障是发生在内网还是外网时，用的命令，包括ping tracert netstat 等

8.求处理机经过并联串联之后的稳定性（就是求它不出现故障的概率）

9.数据库中关于分组查询（group by）并进行排序的SQL语句

其他选择题记不太清了，选择题比较基础，只要掌握了基础知识即可，当然，数据结构最重要

后面就是两个程序填充题，还有一个程序设计题（即附加题）

第一个程序题非常基础，我的印象不深了

第二个程序题大体上就是动态规划的样子。一个城市有N个公交站，每个公交站都有发往其他站的单向车（相当于一个有向图），告诉你每个站发出的车辆所经过的站，让你求出从第一站上车到最后一站需要的最少换乘次数。

附加题是一个手机输入法问题。实现手机按键后显示汉字或词语，并有联想功能，我的思路是带分支限界的广搜，或者是建立本地数据库，对首字母和其他一些关键词组通过散列表实现对其快速匹配

现在还在等一面的结果，煎熬啊，祝你好运噢

第四篇：实习生定位及招聘

北京诚济制药有限公司2014年实习生招聘计划

一、生产中心：

2014年预计招聘实习生25人，具体要求如下：

1、QA:3名（男：2名女：1名）

学历：本科

专业：应用化学、药学、生物工程及药学相关专业

2、QC：4名（男：2名女：2名）

学历：本科

专业：应用化学、药学、生物工程及药学相关专业

3、技术部：3名（男：1名女：2名）

学历：本科

专业：应用化学、药学、生物工程及药学相关专业

4、工程部（设备维修、设备管理）：3人（男）

学历：中专及技校以上

专业：机械设计制造及工程机械相关专业

5、综合办公室：2名（男）

学历：大专以上

专业：工商管理、商务管理、现代企业管理及相关专业

6、生产部：10名（男：6名女：4名）

学历：中专及技校以上

专业：药学相关专业

实习期工作计划及目标：

实习生入厂后首先进行6个月本岗位学习工作，由经验丰富的老员工指导学习实际操作，把在学校学到的理论充分运用到实际操作工作中，争取做到3个月后50%的实习生在实习岗位可以独立完成工作，6个月之内90%以上的实习生可以独立完成本岗位工作，待实习期满后择优录用，对适应工作能力强，表现突出者重点培养，帮助其尽快胜任重要工作岗位。

二、研发中心：

2014年预计招聘实习生6人，具体要求如下：

1、实验员:6名（男女不限）

2、学历：本科

3、专业：分析化学、药学、制药工程及药学相关专业

实习期工作计划及目标：

实习生入厂后首先进行6个月本岗位学习工作，由经验丰富的老员工指导学习实际操作，把在学校学到的理论充分运用到实际操作工作中，争取做到3个月后50%的实习生在实习岗位可以独立完成工作，6个月之内90%以上的实习生可以独立完成本岗位工作，待实习期满后择优录用，对适应工作能力强，表现突出者重点培养，帮助其尽快胜任重要工作岗位。

三、销售中心（普罗）：

市场部

2014年预计招聘实习生4人，具体要求如下：

1、产品专员: 4名（男女不限）

2、学历：大专、本科及以上学历

3、专业：医、药学相关专业

实习期工作计划及目标：

实习生入司后，进行6个月本岗位学习，由市场部产品经理帮带指导，主要进行市场计划的跟进及落实、学术会议的开展、产品培训及市场调研、信息收集等工作。争取在3个月内实现可以独立开展业务，并进行培训等工作。待实习期满后择优录用，并将根据实际情况派到全国各地担任驻地产品专员，对适应工作能力强、表现突出者重点培养。

推广部

2014年预计招聘实习生10人，具体要求如下：

1、招聘岗位：学术专员

2、学历：大专及以上学历

3、专业：医药、营销类相关专业

4、具体工作计划及目标

A、加入公司后集中培训，培训结束后根据培训表现分配至不同的推

广部门；

B、部门经理根据培养计划和工作技能要求及学习进展安排不同的老

员工进行帮带；

C、达到基本工作胜任能力后，安排工作进一步在实际操作中学习；

D、根据工作结果，及时调整培养策略，表现优秀的分配予挑战及机

会更大的项目；

E、工作技能的学习与职业化素养的打造同步进行，每周三及周六定

时举行培训沙龙；

5、职位晋升：

推广系统：学术专员—高级学术专员—teamleader—地区经理—大区经理—推广总监

市场系统：学术专员—高级学术专员—产品专员—产品经理—市场总监

激情成就梦想，拼搏铸造成功！

诚济制药诚邀有志学子共创辉煌！

第五篇：2012微软“智在未来”实习生招聘

2012微软“智在未来”实习生招聘

2012微软“智在未来”实习生计划是为即将毕业于2013年的本科、硕士、博士学生量身打造的实习计划。

 实习生职位：职位涵盖软件开发，技术支持等领域 工作地址：将在北京、上海等城市共提供近1000个实习岗位

“智在未来”实习生计划在暑假期间将招聘大量2013年即将毕业的学生加入微软中国实习。在暑假以外的时间，微软中国也会依据部门需要和同学的时间招聘一定量的实习生。

实习生在微软的实习期通常需要达到三个月或以上，形式分全职兼职两种，全职实习生周一至周五工作5天，兼职实习生一周需要保证工作3个工作日以上。我们鼓励同学们在假期进行全职实习，以获得更全面连贯的实习项目经验.我们能提供什么？

 导师制度：不用担心进入微软后手足无措，我们将为每位实习生配备导师，你的日常学习和工作将由他

（她）悉心指导。他（她）将指引你成功完成整个实习计划。短短几年，从实习生到团队经理，他（她）深谙微软支持工程师所必需的职业素养，帮助你高效完成从校园人才到企业精英的转变；

 培训：为期一周的训练中，你的经理给你制定最适合的培训课程，训练你的沟通能力和职业意识，并在此基础上为你提供可能的职业规划建议；

 技术项目真实现场：每位实习生都将得到自己的项目任务，你将通过团队合作和学习，投身真实项目工

作，并在实习结束时反馈自己的项目报告

你将有何获益？

 提前体验微软 IT 前沿技术：你有机会接触到微软最新产品的测试，领先一步掌握全球 IT 科技资讯；你

有机会与技术大牛们分享交流最新的行业动态和展示微软最新的技术成果；



 获得微软的技术资源：为实习生提供的在线技术自学环境，将让你感受获取知识的乐趣； 绿色通道：表现优秀的实习生将有机会受到微软导师的推荐，经过面试筛选，得到微软2013年正式聘

用书，并于2013年毕业后正式入职，成为微软中国的正式员工。

想加入微软“智在未来”实习生项目吗？亲爱的同学，我们在等你们，那你还等什么呢？赶快来加入微软吧！请访问 申请我们的实习职位。暑期实习生招聘的网申截至日期是3月31日。请同学们火速提交简历，为即将举办的笔试占据你的一席之地！