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104年 高考三級 資訊處理｜歷屆考題全文（機器抽取）

104年公務人員高等考試三級考試暨普通考試資訊處理類科歷屆試題，含國文（作文、公文與測驗）、法學知識與英文（包括中華民國憲法、法學緒論、英文）、資料結構等 8 科。

共 8 份考卷｜資料來源：依政府資料開放授權條款（OGDL）第 1 版利用，資料集：考選部歷屆試題（data.gov.tw dataset 170565），104 年。

本頁文字由 PDF 機器抽取，可能有辨識誤差，僅供搜尋參考，請以官方原始檔案為準。

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國文（作文、公文與測驗）

下載：題目　答案

104 年公務人員高等考試三級考試試題

代號： 20110 | 30010 全一張 （正面）

類 科：

各類科（除公職土木工程技師、公職醫事檢驗師、公職藥師、公職 護理師、公職諮商心理師、公職營養師、公職食品技師外）

科 目：

國文（作文、公文與測驗）

考試時間：

2 小時

座號：

※注意：

禁止使用電子計算器。

甲、作文與公文部分：

一、 作文：（ 60 分）

言論自由是民主社會的基石，所以法律對言論自由給予明文保障，然而針對社會議題的 批評，若查證不實，推論失當，則可能誤導群眾，毀人名譽，產生不良的後果。請以 「言論自由與自律」為題，作文一篇，深入說明你的看法。

二、 公文：（ 20 分）

齊柏林先生拍攝的紀錄影片「看見臺灣」，讓國人驚見臺灣國土之美，但也暴露土地濫 墾、濫伐及河川污染之嚴重，令人怵目驚心，為免引發更大浩劫，亟待設法導正與杜絕。 試擬行政院環境保護署致各直轄市政府、縣市政府函：請加強宣導正確環保觀念，針對轄 區內之土地及河川，建置完善的監測、預警、通報及應變系統，對於違反環保法令事件，應

依法嚴辦，並於三個月內查處完竣，以提昇國人生活品質。

乙、測驗部分：（ 20 分）

代號： 1201

 甲：項羽、乙：張良、丙：韓信、丁：范增

104 年公務人員高等考試三級考試試題

代號：

20110

|

30010

全一張

（背面）

類 科： 各類科（除公職土木工程技師、公職醫事檢驗師、公職藥師、公職 護理師、公職諮商心理師、公職營養師、公職食品技師外）

科 目：

國文（作文、公文與測驗）

依據上文，下列何者所述符合掌握勝利的五種情境： ① 主帥必須精準分辨可開戰與不可開 戰的對象。 ② 主帥必須依據實力來決定戰術。 ③ 主帥應使全軍上下同心奮戰以取勝。 ④ 主帥 應對下屬做有效監督，隨時準備對抗對手。

法學知識與英文（包括中華民國憲法、法學緒論、英文）

下載：題目　答案

代號：

3201

頁次：

4 － 1

104 年公務人員高等考試三級考試試題

類 科：各類科（除公職土木工程技師、公職醫事檢驗師、公職藥師、公職護理師、 公職諮商心理師、公職營養師、公職食品技師外）

科 目：法學知識與英文（包括中華民國憲法、法學緒論、英文）

考試時間：

1 小時

座號：

※注意：

本科目共

50

題，每題

2

分，須用

2B

鉛筆在試卡上依題號清楚劃記，於本試題上作答者，不予計分。

 以行政規則定之  由憲法本身加以規定

 公務人員保障暨培訓委員會

 公務人力發展中心

 第二讀會

 第三讀會

該規定有害及寺廟信仰之傳布存續，對宗教活動自由之限制尚未逾越必要之程度

該規定與憲法第

7

條之宗教平等原則仍屬相符

下列何種考試，不屬憲法第

公務人員高等考試

18

條應考試權之範圍？

外交領事人員考試

專技人員律師考試

 監察院

10

 保姆證照考試

比較沒有爭議的法律案可以直接進入三讀程序

所有法律案皆須經過立法院有關委員會之審查

不動產役權

質權

留置權

抵押權

甲為乙法人之董事，甲執行職務加損害於他人

甲男與未成年之乙女結婚前，乙女因其父母去世，而由甲男擔任監護人，且結婚未逾一年

得減輕

必減輕

 減輕或免除其刑

對於故意或過失不法侵害其著作財產權者，權利人得請求損害賠償

 不罰

請依下文回答第 37 題至第 40 題

An airplane maker, an airline and a biofuel company are working together to make fuel 37 tobacco plant seed oil. The companies are Boeing, South African Airways and SkyNRG. They are using a new tobacco plant 38 'Solaris.' The Dutch biofuel company SkyNRG developed the plant. It contains less of the drug nicotine than traditional tobacco.

Julie Felgar works on environmental issues for Boeing. She said the plant also has many more seeds than traditional tobacco plants 39 . She said only the oil from the seeds will 40 to make biofuel now. But researchers are trying to develop ways to use the entire plant to make fuel.

Ian Cruickshank is an environmental issues specialist for South African Airways Group. He said the special tobacco permits growth of a marketable biofuel crop without supporting smoking.

 glow

 use

請依下文回答第 41 題至第 45 題

Many are the journalists who dream about reporting on the world's deadliest scenes of strife. Few are the ones who actually do it. Even fewer are those who do it well. Camille Lepage, a young photographer from France who was educated in the U.K. but 41 work took her far from Europe, was among the latter.

Last fall, Lepage ventured into the Central Africa Republic, where a coup staged by mainly Muslim rebels had crumpled the state and prompted largely Christian militias to retaliate with fury. She spent her time 42 the fighters and also those most affected by the turmoil, while learning all she could about its roots. William Daniels, a photographer who worked with her here, 43 that locals respected her commitment and professionalism. 'She was very active, very patient, very passionate about this work,' he says. 'Very brave.'

On May 13 the office of French President Francois Hollande issued a statement 44 Lepage's death, at the age of 26, which it described as murder. French peacekeeping troops had discovered her body after they stopped a car driven by antibalaka militiamen. On a recent trip to New York City, Lepage admitted to fears about the conflict she was covering but said that she did not seek out scenes of violence on purpose. 45 , she wanted, in a way typical of her, to find the humanity in that bleak situation. The pictures she made in her brief life will define how she will be remembered, but the pictures she was going to take will help form her legacy.

 that

 to photograph

 photographing

 specializes

 recovering

 supporting

 Otherwise

請依下文回答第 46 題至第 50 題

Las Vegas is a good restaurant town. It offers respectable culinary and ethnic diversity, served dependably. Hotel dining in Las Vegas is relatively homogeneous in style and cuisine, while proprietary restaurants try hard to be different. The restaurant business in Las Vegas is as much a psychological as a culinary art. In Las Vegas you can have the same meal in an astounding variety of environments for an unbelievable range of prices.

Left to its own devices, Las Vegas would be a meat and potatoes town. Owing to the expectations of its many visitors, however, Las Vegas restaurants make things extra special. There are dozens of designer restaurants, gourmet rooms as they are known locally, where the pampered and the curious can pretend they are dining in an exclusive French or Continental restaurant while enjoying the food they like most: meat and potatoes.

There are two kinds of restaurants in Las Vegas: restaurants which are an integral part of a hotel/casino operation, and restaurants which must make it entirely on the merits of their food. Gourmet rooms in the hotels are usually associated with the casinos. Their mission is to pamper customers who are giving the house a lot of gambling action. At any given time, most of the folks in a hotel gourmet room are dining as guests of the casino. If you are paying customers in the same restaurant, the astronomical prices you are charged help subsidize the feeding of all these complimentary guests. Every time you buy a meal in a gourmet room, you are helping to pay the tab of the strangers sitting at the next table. This is not to say the gourmet rooms do not serve excellent food. On the contrary, some of the best chefs in the country cook for hotel/casino gourmet rooms. The bottom line, however, if you are a paying guest, is that you are taking up space intended for high rollers, and the house is going to charge you a lot of rent.

Restaurants independent of casinos work at a considerable disadvantage. First, they do not have a captive audience of gamblers. Second, their operation is not subsidized by gaming, and third, they are not located where you will just stumble upon them. Finally, they not only compete with the casino gourmet rooms, but also go head-to-head with the numerous buffets and bulk-loading meal deals which casinos offer as loss-leaders to attract the less affluent gambler.

資料結構

下載：題目

官方不公布申論答案

類 科： 資訊處理

科 目： 資料結構

考試時間：

2 小時

※注意：

 禁止使用電子計算器。

座號：

if ) ( m l < push ) , ( l m into stack

if ) 1 ( r m < + push ) 1 , ( r m + into stack

一開始，堆疊中只有一組資料， ) ( 1 , n 表示 ] [1.. n A 需要排序。如此反覆將堆疊最上面 的資料 ) , ( r l 移出，執行 partition ) , , ( r l A ，直到堆疊沒有資料為止。

（每小題 10 分，共 20 分）

程式語言

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104 年公務人員高等考試三級考試試題

類 科： 資訊處理

科 目： 程式語言

考試時間：

2 小時

※注意：

 禁止使用電子計算器。

 不必抄題，作答時請將試題題號及答案依照順序寫在試卷上，於本試題上作答者，不予計分。

S → NP VP | VP

NP → ADJ NP | N

VP → ADV VP | V NP | V

N → 考 | 校 | 評鑑 | 成績

V → 通過 | 獲得 | 成功 | 失敗

Adj → 高 | 普 | 特 | 好 | 壞

Adv → 一定 | 可能

繪製 ⑴ 特校可能通過評鑑

相關推導過程的 Parse Tree 。

 修改 TesterID 5 考生的數學成績為 A 。

代號：

26840

（正面）

座號：

類 科： 資訊處理

科 目： 程式語言

SmartCard

-cardID: String

-issueDate: String

-issueOrg: Organization

#getCardID(): String

#getIssueDate(): String

#getIssueOrg(): Organization

#setCardID(String): void

#setIssueDate(String): void

#setIssueOrg(Organization): void

+nowDateToString(): String

+toString():String

Security

SMsecurity

-date: String

-securityList: List

-enterExit: EnterExit

+SMsecurity()

+Security()

+SMsecurity(String, Organization)

+Security(EnterExit)

+getDate(): String

+addInfo(EnterExit): void

+getEnterExit(): EnterExit

+addInfo(String, EnterExit):void

+setData(EnterExit): void

+addInfo(Security):void

+setData(String, EnterExit): void

+toString(): String

+toString(): String

SMsecurity sms = new SMsecurity("SM001", Organization.NewTaipeiMRT); sms.addInfo("2015/06/15 18:36:01", EnterExit.Enter); sms.addInfo("2015/06/15 20:16:01", EnterExit.Exit); sms.addInfo("2015/06/20 18:20:01", EnterExit.Enter); sms.addInfo(EnterExit.Exit); /* for Java */ System.out.println(sms.toString()); /* for C# */ Console.WriteLine(sms.ToString()); 應能產生類似如下的結果

*****Smart card SM001 (6/21/2015 10:38:38 PM, NewTaipeiMRT) -----Enter and Exit information: (2015/06/15 18:36:01, Enter) (2015/06/15 20:16:01, Exit) (2015/06/20 18:20:01, Enter) (6/21/2015 10:38:38 PM, Exit)

代號：

26840

Tester Grade
TesterID TesterID (FK) SubjectID
Name GradeData Subject

資通網路

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官方不公布申論答案

類 科： 資訊處理

科 目： 資通網路

考試時間：

2 小時

※注意：

禁止使用電子計算器。

座號：

系統專案管理

下載：題目

官方不公布申論答案

類 科： 資訊處理

科 目： 系統專案管理

考試時間：

2 小時

※注意：

 禁止使用電子計算器。

四、針對下列八支程式模組：

Cohesion Ratio

=

Number of program modules having functional cohesion

Total number of program modules

（請接第二頁）

代號：

26860

全四頁

第一頁

座號：

cusum += counter; product *= counter; private int cusum, product; //P3

類 科： 資訊處理 科 目： 系統專案管理 //P1 public class P1 { public void count1(int m, int n, int p) { int counter1, counter2, counter3; counter1 = 1; cusum = 0; while (counter1 <= m) { cusum += counter1; counter1 += 1; } counter2 = 1; product = 1; while (counter2 <= n) { product *= counter2; counter2 += 1; } counter3 = 1; sum = 0; while (counter3 <= p) { sum += counter3; counter3 += 1; } mean = sum / p; } public int getSum() { return sum; } public int getProduct() { return product; } public int getCusum() { return cusum; } public int getMean() { return mean; } private int sum, product, cusum, mean; } //P2 public class P2 { public void count2(int n) { int counter; counter = 1; cusum = 0; product = 1; while (counter <= n) { counter += 1; } } public int getCusum() { return cusum; } public int getProduct() { return product; } }

cusum += counter;

public class P3 { public void count3(int n) { int counter; counter = 1; cusum = 0; while (counter <= n) { counter += 1; } mean = cusum / n; } public int getCusum() { return cusum; } public int getMean() { return mean; } private int cusum, mean; }

類 科： 資訊處理

科 目： 系統專案管理 //P4 public class P4 { public void count4(int n) { int counter; counter = 1; cusum = 0; while (counter <= n) { cusum += counter; counter += 1; } } public int getCusum() { return cusum; } private int cusum; } //P5 public class P5 { public void count5(int first,int second) { int intermediate; intermediate = first; result_first = second; result_second = intermediate; } public int getResult_first() { return result_first; } public int getResult_second() { return result_second; } private int result_first, result_second; } //P6 public class P6 { public void count6(int n,int product) { int counter1, counter2, counter3, counter4; counter1 = 1; int a[] = new int[n]; while (counter1 <= n) { a[counter1-1] = counter1; counter1 += 1; } counter2 = 0; cusum = 0; while (counter2 < n) { cusum += a[counter2]; counter2 += 1; } counter3 = 0; prod = 1; while (counter3 < n) { prod = prod* product * a[counter3]; counter3 += 1; } counter4 = 0; sum = 0; while (counter4 < n) { sum += a[counter4]; counter4 += 1; } mean = sum / n; } public int getCusum() { return cusum; } public int getProd() { return prod; } public int getSum() { return sum; } public int getMean() { return mean; } private int cusum, prod, sum, mean;

}

類 科： 資訊處理

科 目： 系統專案管理

//P7 public class P7 { public void count7(int[] tmp, int n) { int counter1, counter2, temp; counter1 = 0; a = tmp; System.out.print("\n"); for (counter1 = 1; counter1 < n; counter1++) { for (counter2= 0; counter2 < counter1; counter2++) { if (a[counter1] <a[counter2]) { temp = a[counter1]; a[counter1] = a[counter2]; a[counter2] = temp; } } } } public int[] geta() { return a; } private int[] a; } //P8 public class P8 { public void count8(int m, int n, int p, int flag) { int counter1, counter2, counter3; cusum = 0; product = 1; sum = 0; mean = 0; if (flag == 1) { counter1 = 1; cusum = 0; while (counter1 <= m) { cusum += counter1; counter1 += 1; } } else if (flag == 2) { counter2 = 1; product = 1; while (counter2 <= n) { product *= counter2; counter2 += 1; } } else { counter3 = 1; sum = 0; while (counter3 <= p) { sum += counter3; counter3 += 1; } } mean = sum / p; } public int getCusum() { return cusum; } public int getProduct() { return product; } public int getSum() { return sum; } public int getMean() { return mean; } private int cusum, product, sum, mean; }

代號：

26860

全四頁

第四頁

內聚力型態 所對應之程式模組 （請以 P1, P2,… 等標示） 說明
…… …… ……

資料庫應用

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官方不公布申論答案

104 年公務人員高等考試三級考試試題

類 科： 資訊處理

科 目： 資料庫應用

考試時間：

2 小時

※注意：

 禁止使用電子計算器。

Sailors(sid:integer, sname:string, rating:integer, age:real)

Boats(bid: integer, bname:string, color:string)

Reserves(sid: integer, bid:integer, day:date)

圖一、 Sailors 表格的案例

圖三、 Boats 表格的案例

請回答下列各題：（每小題 5 分，共 40 分）。請注意：答案必須具備一般性，表 格案例內容只為參考用，案例內容更改後，答案仍須正確。

請劃出此資料庫相對應的實體關聯圖（

ER Diagram

）。

代號：

26870

（正面）

座號：

類 科： 資訊處理

科 目： 資料庫應用

A

→ BC

CD

B

E

→ A

請問將 r 切割為 r1(A,B,C) 及 r2(A,D,E) 的切割是否是無損切割？若是無損切割，則 請證明之。（ 10 分）

圖四、交易 T1 、 T2 的執行過程（指令的上下位置表示執行時間的先後，上面的指令比下面的 指令先發生）

→ D

→ E

代號：

sid sname rating age
22 Wawrinka 10 30
29 Brutus 6 33
31 Lubber 8 55
32 Lu 9 32
58 Rusty 8 35
64 Nishikori 10 26
71 Dustin 10 30
74 Murray 10 28
85 Nadal 10 29
95 Bob 3 60
bid bname color
101 Intelake blue
102 Clipper red
103 Marine green
104 Blast red
sid bid day
22 101 6/7/2015
22 102 6/5/2015
22 103 6/3/2015
22 104 6/1/2015
32 102 5/28/2015
32 103 6/1/2015
64 104 6/3/2015
85 101 5/30/2015
85 102 6/3/2015
85 103 6/3/2015
T1 T2
Read(A)
A:=A-50
Write(A)
Read(B)
B:=B-10
Write(B)
Read(A)
A:=A+10
Write(A)
Read(B)
B:=B+50
Write(B)

資訊管理與資通安全

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官方不公布申論答案

類 科： 資訊處理

科 目： 資訊管理與資通安全

考試時間：

2 小時

座號：

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