第一篇：微軟2014實(shí)習(xí)生招聘及技術(shù)類秋令營(yíng)招聘onlinetest編程題

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(chǎn)’ and ‘z’(including ‘0’, ‘9’, ‘a(chǎn)’, ‘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(chǎn)’ is larger than ‘9’, and ‘z’ is larger than ‘a(chǎn)’(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：微軟2014實(shí)習(xí)生招聘及技術(shù)類秋令營(yíng)招聘onlinetest編程題

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(chǎn)’ and ‘z’(including ‘0’, ‘9’, ‘a(chǎn)’, ‘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(chǎn)’ is larger than ‘9’, and ‘z’ is larger than ‘a(chǎn)’(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：騰訊2010實(shí)習(xí)生招聘筆試題目(技術(shù)類)

1.第一題是軟件工程中關(guān)于聚合在UML中的畫法

空心菱靠在整體類上

2.有一個(gè)題是求一個(gè)圖中最小生成樹的權(quán)值總和

3.有一個(gè)是根據(jù)后序和中序求樹的先序遍歷

一個(gè)是關(guān)于sizeof的題目，char A[]=“hello”，求sizeof(A),這一問答案是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.有一個(gè)題目是求散列表存儲(chǔ)數(shù)據(jù)之后的數(shù)組里面各單元存儲(chǔ)的數(shù)據(jù)內(nèi)容

5.有一個(gè)是判斷關(guān)于RISC（精簡(jiǎn)指令）的特點(diǎn)

6.有一個(gè)是選擇進(jìn)程的執(zhí)行順序防止死鎖

7.當(dāng)判斷網(wǎng)絡(luò)故障是發(fā)生在內(nèi)網(wǎng)還是外網(wǎng)時(shí)，用的命令，包括ping tracert netstat 等

8.求處理機(jī)經(jīng)過并聯(lián)串聯(lián)之后的穩(wěn)定性（就是求它不出現(xiàn)故障的概率）

9.數(shù)據(jù)庫中關(guān)于分組查詢（group by）并進(jìn)行排序的SQL語句

其他選擇題記不太清了，選擇題比較基礎(chǔ)，只要掌握了基礎(chǔ)知識(shí)即可，當(dāng)然，數(shù)據(jù)結(jié)構(gòu)最重要

后面就是兩個(gè)程序填充題，還有一個(gè)程序設(shè)計(jì)題（即附加題）

第一個(gè)程序題非?；A(chǔ)，我的印象不深了

第二個(gè)程序題大體上就是動(dòng)態(tài)規(guī)劃的樣子。一個(gè)城市有N個(gè)公交站，每個(gè)公交站都有發(fā)往其他站的單向車（相當(dāng)于一個(gè)有向圖），告訴你每個(gè)站發(fā)出的車輛所經(jīng)過的站，讓你求出從第一站上車到最后一站需要的最少換乘次數(shù)。

附加題是一個(gè)手機(jī)輸入法問題。實(shí)現(xiàn)手機(jī)按鍵后顯示漢字或詞語，并有聯(lián)想功能，我的思路是帶分支限界的廣搜，或者是建立本地?cái)?shù)據(jù)庫，對(duì)首字母和其他一些關(guān)鍵詞組通過散列表實(shí)現(xiàn)對(duì)其快速匹配

現(xiàn)在還在等一面的結(jié)果，煎熬啊，祝你好運(yùn)噢

第四篇：實(shí)習(xí)生定位及招聘

北京誠濟(jì)制藥有限公司2014年實(shí)習(xí)生招聘計(jì)劃

一、生產(chǎn)中心：

2014年預(yù)計(jì)招聘實(shí)習(xí)生25人，具體要求如下：

1、QA:3名（男：2名女：1名）

學(xué)歷：本科

專業(yè)：應(yīng)用化學(xué)、藥學(xué)、生物工程及藥學(xué)相關(guān)專業(yè)

2、QC：4名（男：2名女：2名）

學(xué)歷：本科

專業(yè)：應(yīng)用化學(xué)、藥學(xué)、生物工程及藥學(xué)相關(guān)專業(yè)

3、技術(shù)部：3名（男：1名女：2名）

學(xué)歷：本科

專業(yè)：應(yīng)用化學(xué)、藥學(xué)、生物工程及藥學(xué)相關(guān)專業(yè)

4、工程部（設(shè)備維修、設(shè)備管理）：3人（男）

學(xué)歷：中專及技校以上

專業(yè)：機(jī)械設(shè)計(jì)制造及工程機(jī)械相關(guān)專業(yè)

5、綜合辦公室：2名（男）

學(xué)歷：大專以上

專業(yè)：工商管理、商務(wù)管理、現(xiàn)代企業(yè)管理及相關(guān)專業(yè)

6、生產(chǎn)部：10名（男：6名女：4名）

學(xué)歷：中專及技校以上

專業(yè)：藥學(xué)相關(guān)專業(yè)

實(shí)習(xí)期工作計(jì)劃及目標(biāo)：

實(shí)習(xí)生入廠后首先進(jìn)行6個(gè)月本崗位學(xué)習(xí)工作，由經(jīng)驗(yàn)豐富的老員工指導(dǎo)學(xué)習(xí)實(shí)際操作，把在學(xué)校學(xué)到的理論充分運(yùn)用到實(shí)際操作工作中，爭(zhēng)取做到3個(gè)月后50%的實(shí)習(xí)生在實(shí)習(xí)崗位可以獨(dú)立完成工作，6個(gè)月之內(nèi)90%以上的實(shí)習(xí)生可以獨(dú)立完成本崗位工作，待實(shí)習(xí)期滿后擇優(yōu)錄用，對(duì)適應(yīng)工作能力強(qiáng)，表現(xiàn)突出者重點(diǎn)培養(yǎng)，幫助其盡快勝任重要工作崗位。

二、研發(fā)中心：

2014年預(yù)計(jì)招聘實(shí)習(xí)生6人，具體要求如下：

1、實(shí)驗(yàn)員:6名（男女不限）

2、學(xué)歷：本科

3、專業(yè)：分析化學(xué)、藥學(xué)、制藥工程及藥學(xué)相關(guān)專業(yè)

實(shí)習(xí)期工作計(jì)劃及目標(biāo)：

實(shí)習(xí)生入廠后首先進(jìn)行6個(gè)月本崗位學(xué)習(xí)工作，由經(jīng)驗(yàn)豐富的老員工指導(dǎo)學(xué)習(xí)實(shí)際操作，把在學(xué)校學(xué)到的理論充分運(yùn)用到實(shí)際操作工作中，爭(zhēng)取做到3個(gè)月后50%的實(shí)習(xí)生在實(shí)習(xí)崗位可以獨(dú)立完成工作，6個(gè)月之內(nèi)90%以上的實(shí)習(xí)生可以獨(dú)立完成本崗位工作，待實(shí)習(xí)期滿后擇優(yōu)錄用，對(duì)適應(yīng)工作能力強(qiáng)，表現(xiàn)突出者重點(diǎn)培養(yǎng)，幫助其盡快勝任重要工作崗位。

三、銷售中心（普羅）：

市場(chǎng)部

2014年預(yù)計(jì)招聘實(shí)習(xí)生4人，具體要求如下：

1、產(chǎn)品專員: 4名（男女不限）

2、學(xué)歷：大專、本科及以上學(xué)歷

3、專業(yè)：醫(yī)、藥學(xué)相關(guān)專業(yè)

實(shí)習(xí)期工作計(jì)劃及目標(biāo)：

實(shí)習(xí)生入司后，進(jìn)行6個(gè)月本崗位學(xué)習(xí)，由市場(chǎng)部產(chǎn)品經(jīng)理幫帶指導(dǎo)，主要進(jìn)行市場(chǎng)計(jì)劃的跟進(jìn)及落實(shí)、學(xué)術(shù)會(huì)議的開展、產(chǎn)品培訓(xùn)及市場(chǎng)調(diào)研、信息收集等工作。爭(zhēng)取在3個(gè)月內(nèi)實(shí)現(xiàn)可以獨(dú)立開展業(yè)務(wù)，并進(jìn)行培訓(xùn)等工作。待實(shí)習(xí)期滿后擇優(yōu)錄用，并將根據(jù)實(shí)際情況派到全國(guó)各地?fù)?dān)任駐地產(chǎn)品專員，對(duì)適應(yīng)工作能力強(qiáng)、表現(xiàn)突出者重點(diǎn)培養(yǎng)。

推廣部

2014年預(yù)計(jì)招聘實(shí)習(xí)生10人，具體要求如下：

1、招聘崗位：學(xué)術(shù)專員

2、學(xué)歷：大專及以上學(xué)歷

3、專業(yè)：醫(yī)藥、營(yíng)銷類相關(guān)專業(yè)

4、具體工作計(jì)劃及目標(biāo)

A、加入公司后集中培訓(xùn)，培訓(xùn)結(jié)束后根據(jù)培訓(xùn)表現(xiàn)分配至不同的推

廣部門；

B、部門經(jīng)理根據(jù)培養(yǎng)計(jì)劃和工作技能要求及學(xué)習(xí)進(jìn)展安排不同的老

員工進(jìn)行幫帶；

C、達(dá)到基本工作勝任能力后，安排工作進(jìn)一步在實(shí)際操作中學(xué)習(xí)；

D、根據(jù)工作結(jié)果，及時(shí)調(diào)整培養(yǎng)策略，表現(xiàn)優(yōu)秀的分配予挑戰(zhàn)及機(jī)

會(huì)更大的項(xiàng)目；

E、工作技能的學(xué)習(xí)與職業(yè)化素養(yǎng)的打造同步進(jìn)行，每周三及周六定

時(shí)舉行培訓(xùn)沙龍；

5、職位晉升：

推廣系統(tǒng)：學(xué)術(shù)專員—高級(jí)學(xué)術(shù)專員—teamleader—地區(qū)經(jīng)理—大區(qū)經(jīng)理—推廣總監(jiān)

市場(chǎng)系統(tǒng)：學(xué)術(shù)專員—高級(jí)學(xué)術(shù)專員—產(chǎn)品專員—產(chǎn)品經(jīng)理—市場(chǎng)總監(jiān)

激情成就夢(mèng)想，拼搏鑄造成功！

誠濟(jì)制藥誠邀有志學(xué)子共創(chuàng)輝煌！

第五篇：2012微軟“智在未來”實(shí)習(xí)生招聘

2012微軟“智在未來”實(shí)習(xí)生招聘

2012微軟“智在未來”實(shí)習(xí)生計(jì)劃是為即將畢業(yè)于2013年的本科、碩士、博士學(xué)生量身打造的實(shí)習(xí)計(jì)劃。?

? 實(shí)習(xí)生職位：職位涵蓋軟件開發(fā)，技術(shù)支持等領(lǐng)域 工作地址：將在北京、上海等城市共提供近1000個(gè)實(shí)習(xí)崗位

“智在未來”實(shí)習(xí)生計(jì)劃在暑假期間將招聘大量2013年即將畢業(yè)的學(xué)生加入微軟中國(guó)實(shí)習(xí)。在暑假以外的時(shí)間，微軟中國(guó)也會(huì)依據(jù)部門需要和同學(xué)的時(shí)間招聘一定量的實(shí)習(xí)生。

實(shí)習(xí)生在微軟的實(shí)習(xí)期通常需要達(dá)到三個(gè)月或以上，形式分全職兼職兩種，全職實(shí)習(xí)生周一至周五工作5天，兼職實(shí)習(xí)生一周需要保證工作3個(gè)工作日以上。我們鼓勵(lì)同學(xué)們?cè)诩倨谶M(jìn)行全職實(shí)習(xí)，以獲得更全面連貫的實(shí)習(xí)項(xiàng)目經(jīng)驗(yàn).我們能提供什么？

? 導(dǎo)師制度：不用擔(dān)心進(jìn)入微軟后手足無措，我們將為每位實(shí)習(xí)生配備導(dǎo)師，你的日常學(xué)習(xí)和工作將由他

（她）悉心指導(dǎo)。他（她）將指引你成功完成整個(gè)實(shí)習(xí)計(jì)劃。短短幾年，從實(shí)習(xí)生到團(tuán)隊(duì)經(jīng)理，他（她）深諳微軟支持工程師所必需的職業(yè)素養(yǎng)，幫助你高效完成從校園人才到企業(yè)精英的轉(zhuǎn)變；

? 培訓(xùn)：為期一周的訓(xùn)練中，你的經(jīng)理給你制定最適合的培訓(xùn)課程，訓(xùn)練你的溝通能力和職業(yè)意識(shí)，并在此基礎(chǔ)上為你提供可能的職業(yè)規(guī)劃建議；

? 技術(shù)項(xiàng)目真實(shí)現(xiàn)場(chǎng)：每位實(shí)習(xí)生都將得到自己的項(xiàng)目任務(wù)，你將通過團(tuán)隊(duì)合作和學(xué)習(xí)，投身真實(shí)項(xiàng)目工

作，并在實(shí)習(xí)結(jié)束時(shí)反饋?zhàn)约旱捻?xiàng)目報(bào)告

你將有何獲益？

? 提前體驗(yàn)微軟 IT 前沿技術(shù)：你有機(jī)會(huì)接觸到微軟最新產(chǎn)品的測(cè)試，領(lǐng)先一步掌握全球 IT 科技資訊；你

有機(jī)會(huì)與技術(shù)大牛們分享交流最新的行業(yè)動(dòng)態(tài)和展示微軟最新的技術(shù)成果；

?

? 獲得微軟的技術(shù)資源：為實(shí)習(xí)生提供的在線技術(shù)自學(xué)環(huán)境，將讓你感受獲取知識(shí)的樂趣； 綠色通道：表現(xiàn)優(yōu)秀的實(shí)習(xí)生將有機(jī)會(huì)受到微軟導(dǎo)師的推薦，經(jīng)過面試篩選，得到微軟2013年正式聘

用書，并于2013年畢業(yè)后正式入職，成為微軟中國(guó)的正式員工。

想加入微軟“智在未來”實(shí)習(xí)生項(xiàng)目嗎？親愛的同學(xué)，我們?cè)诘饶銈儯悄氵€等什么呢？趕快來加入微軟吧！請(qǐng)?jiān)L問 申請(qǐng)我們的實(shí)習(xí)職位。暑期實(shí)習(xí)生招聘的網(wǎng)申截至日期是3月31日。請(qǐng)同學(xué)們火速提交簡(jiǎn)歷，為即將舉辦的筆試占據(jù)你的一席之地！