QUESTION 2Identify the memory component from which memory is allocated for:session memory for the shared server, buffers for I/O Slaves Oracle Database Recovery Manager (RMAN) Backup and restore operationsA. Large PoolB. Redo Log BufferC. Database Buffer CacheD. Program Global Area (PGA)Correct answer:a
Explanation/reference:Large PoolThe large pool is a optional memory area intended for memory allocations, is larger than was appropriate for the Shar Ed Pool. The large pool can prov
#include #include using namespace Std;struct nodeDefine Structure body{int num;//DataCount of occurrences of int count;//};BOOL CMP (Node A, node B)//Sort{if (a.count! = b.count)Return a.count>b.count;Return a.num}Node a[1005];Node temp;int main (){int n,x,i;for (i = 0; I {A[i].num = i;}CIN >> N;for (i = 1; I {CIN >> X;++a[x].count;}Sort (A, a + 1005, CMP);for (i = 0; I {if (A[i].count = = 0)Breakcout }return 0;}CCF Exam
, the data we have given guarantees that the sales of two consecutive days in this N day are always different, namely Ai-1≠ai. Note that if the two days are not contiguous, the sales volume may be the same. Input format the first line of input contains an integer n.The second line contains n integers, separated by spaces, representing A1, A2, ..., an. Output format outputs an integer that represents the number of vertices that appear. Sample Input 75 4 1 2 3 6 4 Sample Output 2 evaluation use ca
that all submitted data is written to the data file during a consistent shutdownInstance recovery time:Refers to the time required to push the last checkpoint (checkpoint location) of the data file to the latest SCN recorded in the control file. Administrators can control the time by setting the mttr target and resizing the redo log group.MTTR:Mean time to Repair, that is, the mean duration of repair. Refers to the average repair time for repairable products, which is the time between failure a
free space in the data block, the new state of the block is reflected in the bitmap. By using bitmaps, the Oracle DB You can increase the degree of automation in managing free space. Therefore, this type of space management is called "Automatic segment Space Management (ASSM)".Manual: This method specifies that you want to use an idle list to manage free space in a segment. A free list is a list of blocks of data that have space to insert rows. Because of this way of managing the space, you nee
Tags: page jump mysql 5.6 base technology share upload TPS ATI Scholarship OptionsMySQL OCP OverviewThe Oracle OCP (Oracle Certified Professional) exam may be familiar to you all. However, since MySQL was acquired by Oracle, Oracle has launched the MySQL OCP exam. Unfortunat
Tags: mysq ima IPC Others different show win service generation Problem environment, recently prepared MySQL exam, with the * * Education question Bank, do the operation of the problem when the database file to copy the data directory, can display the table, but cannot manipulate the table 1 View the directory where MySQL holds the database files 2 Copy the dat
where Setudent.sno in (select Selclass.sno from Selclass GROUP by SELCLASS.SNO have count (*) > = 3);5. Enquiry for students who have enrolled in all coursesSELECT * from Setudent where Setudent.sno in (select Selclass.sno from Selclass GROUP by SELCLASS.SNO have count (*) = (SE Lect Count (*) from coures));6. Number of students inquiring for elective coursesSelect Cno,count (*) from Selclass GROUP by SELCLASS.CNO;7. Search for a course at least one of the same student information as the Jiang
A simple PHP tutorial MySQL tutorial exam system, the example method is to determine whether the topic array is empty, empty then from the question bank randomly removed 12 questions ID
Include './inc/header.php ';if ($_session[' arr '] = = ') {#判断题目数组是否为空, empty then randomly take out 12 questions ID from the question bank$sql = ' select * from ';$select = $db->select ($sql);for ($i =0; $i $aid. = $
score;
INSERT into student (sname, Gender, class_id) VALUES ("Harry", "Male", 3);
INSERT into score (student_id, course_id, number) VALUES (8,5,93); SELECT student.sid number, student.sname name from score INNER JOIN student on score.student_id = Student.sid and Course_
id=5; # 9, check the course number "2" of the grade than the course number "1" of all the students of the lower class number, name; INSERT into student (sname, Gender, class_id) VALUES("Zhao Liu", "female", 2);
INSERT into sco
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