栈的特性是后进先出又称为LIFO表，那么后进先出的英文是（）。A、First In Last OutB、First In First OutC、Last In Last OutD、Last In First Out
栈的特性是后进先出又称为LIFO表，那么后进先出的英文是（）。
- A、First In Last Out
- B、First In First Out
- C、Last In Last Out
- D、Last In First Out
相关考题：
3．In China，the________ is behind________A． first name ;last nameB． last name; family nameC． family name; first nameD． last name; first name
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One attribute of the stack is(72).A．FIFO( First In First Out)B．LIFO( Last In First Out)C．queueD．built into their circuitry
Examine the structure of the EMPLOYEES table:Column name Data type RemarksEMPLOYEE_ID NUMBER NOT NULL, Primary KeyLAST_NAME VARCNAR2(30)FIRST_NAME VARCNAR2(30)JOB_ID NUMBERSAL NUMBERMGR_ID NUMBER References EMPLOYEE_ID column DEPARTMENT_ID NUMBERYou need to create an index called NAME_IDX on the first name and last name fields of the EMPLOYEES table. Which SQL statement would you use to perform this task? （）A. CREATE INDEX NAME _IDX (first_name, last_name);B. CREATE INDEX NAME _IDX (first_name, AND last_name)C. CREATE INDEX NAME_IDX ON (First_name, last_name);D. CREATE INDEX NAME_IDX ON employees (First_name, AND last_name);E. CREATE INDEX NAME_IDX ON employees (First_name, last_name);F. CREATE INDEX NAME_IDX FOR employees (First_name, last_name);
在SQL Server 2000中，现要在employees表的first_name和last_name列上建立一个唯一的非聚集复合索引，其中first_name列数据的重复率是5%，last_name列数据的重复率是10%。请补全下列语句使以first_name和last_name列作为条件的查询效率最高。CREATE UNIQUE NONCLUSTERED INDEX Idx_NameON employees(______)
Given the following query:SELECT last_name, first_name, age, hire_date FROM employee WHERE age >40Which of the following clauses must be added to return the rows sorted by AGE, oldest first, and by LAST_NAME, from A to Z?（）A.SORT BY age ASC, last_nameB.SORT BY age DESC, last_nameC.ORDER BY age DESC, last_nameD.ORDER BY age ASC, last_name
阅读下列说明和C代码，回答问题 1 至问题 3，将解答写在答题纸的对应栏内。 【说明】 假币问题：有n枚硬币，其中有一枚是假币，己知假币的重量较轻。现只有一个天平，要求用尽量少的比较次数找出这枚假币。 【分析问题】 将n枚硬币分成相等的两部分: (1)当n为偶数时，将前后两部分，即 1...n/2和n/2+1...0，放在天平的两端，较轻的一端里有假币，继续在较轻的这部分硬币中用同样的方法找出假币: (2)当n为奇数时，将前后两部分，即1..(n -1)/2和(n+1)/2+1...0，放在天平的两端，较轻的一端里有假币，继续在较轻的这部分硬币中用同样的方法找出假币；若两端重量相等，则中间的硬币，即第 (n+1)/2枚硬币是假币。 【C代码】 下面是算法的C语言实现，其中: coins[]： 硬币数组 first，last：当前考虑的硬币数组中的第一个和最后一个下标 include stdio.h int getCounterfeitCoin(int coins[]， int first，int last) { int firstSum = 0，lastSum = 0; int ; If(first==last-1){ /*只剩两枚硬币*/ if(coins[first] coins[last]) return first; return last; } if((last - first + 1) % 2 ==0){ /*偶数枚硬币*/ for(i = first;i ( 1 );i++){ firstSum+= coins[i]; } for(i=first + (last-first) / 2 + 1;i last +1;i++){ lastSum += coins[i]; } if( 2 ){ Return getCounterfeitCoin(coins,first,first+(last-first)/2;) }else{ Return getCounterfeitCoin(coins,first+(last-first)/2+1,last;) } } else{ /*奇数枚硬币*/ For(i=first;ifirst+(last-first)/2;i++){ firstSum+=coins[i]; } For(i=first+(last-first)/2+1;ilast+1;i++){ lastSum+=coins[i]; } If(firstSumlastSum){ return getCounterfeitCoin(coins,first,first+(last-first)/2-1); }else if(firstSumlastSum){ return getCounterfeitCoin(coins,first+(last-first)/2-1,last); }else{ Return( 3 ) } } }【问题一】 根据题干说明，填充C代码中的空（1）-（3） 【问题二】 根据题干说明和C代码，算法采用了（ ）设计策略。 函数getCounterfeitCoin的时间复杂度为（ ）（用O表示）。 【问题三】 若输入的硬币数为30，则最少的比较次数为（ ），最多的比较次数为（ ）。
The way in which people address each other depends on their age, sex, social group and personal relationship. The English system of address forms frequently used includes first name, last name, title+last name, （）, and kin term.Atitle+first nameBtitle+titleCtitle aloneDfirst name+last name+title
The way in which people address each other depends on their age, sex, social group and personal relationship. The English system of address forms frequently used includes first name, last name, title+last name, （）, and kin term.A、title+first nameB、title+titleC、title aloneD、first name+last name+title
您要对EMPLOYEES表的FIRST_NAME和LAST_NAME列创建一个组合索引。以下哪条语句将完成此任务（）A、CREATE INDEXfl_idx ON employees（first_name last_name）B、CREATE INDEXfl_idx ON employees（first_name），employees（last_name）C、CREATE INDEXfl_idx ON employees（first_name，last_name）D、CREATE INDEXfl_idx ON employees（first_name）；CREATE INDEXfl_idx ON employees（last_name）
下面的文件正确吗？为什么？用IE和本章中的解析器验证你的结论。 mes.dtd ＜!ELEMENT message ANY＞ ＜!ELEMENT persion (name,age?,lxfs)＞ ＜!ELEMENT lxfs (#PCDATA,tel|email)*＞ ＜!ELEMENT name (first,last)＞ ＜!ATTLIST name sex (male|female) "male"＞ ＜!ELEMENT first %pc; ＞ ＜!ELEMENT last %pc;＞ ＜!ELEMENT age %pc;＞ ＜!ELEMENT tel %pc;＞ ＜!ELEMENT email %pc;＞ ＜!ENTITY % pc “(#PCDATA)”＞ ＜!ELEMENT emergency EMPTY＞ ＜!ATTLIST emergency fire CDATA #FIXED "119" police CDATA #FIXED "110" hospital CDATA #FIXED "120" ＞ mes.xml ＜?xml version="1.0" encoding="gb2312"?＞ ＜!DOCTYPE SYSTEM "mes.dtd"＞ ＜message＞ ＜persion＞ ＜name sex="男"＞ ＜first＞li＜/first＞ ＜last＞xiao＜/last＞ ＜/name＞ ＜age＞25＜/age＞ ＜lxfs＞ ＜tel＞123456＜/tel＞ ＜/lxfs＞ ＜/persion＞ ＜persion＞ ＜name＞ ＜first＞wang＜/first＞ ＜last＞xiao＜/last＞ ＜/name＞ ＜lxfs＞＜/lxfs＞ ＜/persion＞ ＜emergency fire="120" police="110" /＞ ＜/message＞
Examine the structure of the EMPLOYEES table: Column name Data type Remarks EMPLOYEE_ID NUMBER NOT NULL, Primary Key LAST_NAME VARCNAR2(30) FIRST_NAME VARCNAR2(30) JOB_ID NUMBER SAL NUMBER MGR_ID NUMBER References EMPLOYEE_ID column DEPARTMENT_ID NUMBER You need to create an index called NAME_IDX on the first name and last name fields of the EMPLOYEES table. Which SQL statement would you use to perform this task? （）A、CREATE INDEX NAME _IDX (first_name, last_name);B、CREATE INDEX NAME _IDX (first_name, AND last_name)C、CREATE INDEX NAME_IDX ON (First_name, last_name);D、CREATE INDEX NAME_IDX ON employees (First_name, AND last_name);E、CREATE INDEX NAME_IDX ON employees (First_name, last_name);F、CREATE INDEX NAME_IDX FOR employees (First_name, last_name);
Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) HIRE_DATE DATE Which UPDATE statement is valid?（）A、UPDATE employees SET first_name = 'John' SET last_name ='Smith' WHERE employee_id = 180;B、UPDATE employees SET first_name = 'John', SET last_name ='Smith' WHERE employee_id = 180;C、UPDATE employees SET first_name = 'John' AND last_name ='Smith' WHERE employee_id = 180;D、UPDATE employees SET first_name = 'John', last_name ='Smith' WHERE employee_id = 180;
单选题The way in which people address each other depends on their age, sex, social group and personal relationship. The English system of address forms frequently used includes first name, last name, title+last name, （）, and kin term.Atitle+first nameBtitle+titleCtitle aloneDfirst name+last name+title
单选题Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2 (25) LAST_NAME VARCHAR2 (25) HIRE_DATE DATE Which UPDATE statement is valid? （）AUPDATE employees SET first_name = 'John' SET last_name = 'Smith' WHERE employee_id = 180;BUPDATE employees SET first_name = 'John', SET last_name = 'Smoth' WHERE employee_id = 180;CUPDATE employee SET first_name = 'John' AND last_name = 'Smith' WHERE employee_id = 180;DUPDATE employee SET first_name = 'John', last_name = 'Smith' WHERE employee_id = 180;
单选题栈的特性是后进先出又称为LIFO表，那么后进先出的英文是（）。AFirst In Last OutBFirst In First OutCLast In Last OutDLast In First Out
多选题Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) Which three statements inserts a row into the table? （）AINSERT INTO employees VALUES (NULL, 'JOHN','Smith');BINSERT INTO employees( first_name, last_name) VALUES ('JOHN','Smith');CINSERT INTO employees VALUES ('1000','JOHN','NULL');DINSERT INTO employees(first_name,last_name, employee_id) VALUES ('1000, 'john','Smith');EINSERT INTO employees (employee_id) VALUES (1000);FINSERT INTO employees (employee_id, first_name, last_name) VALUES ( 1000, 'john',);