（C）1、Suppose that a selection sort of 80 items has completed 32 iterations of the main loop. How many items are now guaranteed to be in their final spot (never to be moved again)?
A、16 B、31
（ABC？）2、Which synchronization mechanism(s) is/are used to avoid race conditions among processes/threads in operating system?
哪个同步机制(s)是/是用来避免操作系统中的进程/线程的竞态条件?
竞态条件即是同步问题
A、Mutex B、Mailbox C、Semaphore D、Local procedure call
A、互斥信号量
（AC）3、There is a sequence of n numbers 1,2,3,...,n and a stack which can keep m numbers at most. push the n numbers into the stack following the sequence and pop out randomly . Suppose n is 2 and m is 3,the output sequence may be 1,2 or 2,1,so we get 2 different sequences . Suppose n is 7,and m is 5,please choose the output sequence of the stack.
A、1,2,3,4,5,6,7
B、7,6,5,4,3,2,1
C、5,6,4,3,7,2,1
D、1,7,6,5,4,3,2
E、3,2,1,7,5,6,4
（A）4、Which is the result of binary number 01011001 after multiplying by 0111001 and adding 1101110?
可先将二进制数字转换为十进制进行运算，算的结果为：5183，将结果装换为十六进制，则5183的十六制最低位为15 ，故结果为A
A、0001010000111111
B、0101011101110011
C、0011010000110101
（C）5、What is output if you compile and execute the following c code?
{
int i = 11;
int const *p = &i;
p++;
printf("%d", *p);
}
产生的应该是i的地址，所以为一个垃圾值
用codeblock运行后显示结果为：
2686748
Process returned 0 (0x0)
Press any key to continue.
(A) 11 (B) 12 (C) Garbage value (D) Compile error (E) None of above
（ABC？）6. Which of following C++ code is correct?
(A) int f()
{
int *a = new int(3);
return *a;
}
(B) int *f()
{
int a[3] = {1, 2, 3};
return a;
}
(C) vector f()
{
vector v(3);
return v;
}
(D) void f(int *ret)
{
int a[3] = {1, 2, 3};
ret = a;
return;
}
（D）7. Given that the 180-degree rotated image of a 5-digit number is another 5-digit number and the difference between the numbers is 78633, what is the original 5-digit number?
将数字旋转180度，和回文数稍有不同，应该是一简单题。
(A) 60918 (B) 91086 (C) 18609 (D) 10968 (E) 86901
（ACD）8. Which of the following statements are true?
(A) We can create a binary tree from given inorder and preorder traversal sequences.
由前、后序遍历序列数可知根节点，故中序和前序可以确定一棵二叉树
同理，B错
(B) We can create a binary tree from given preorder and postorder traversal sequences.
(C) For an almost sorted array, insertion sort can be more effective than Quicksort.
时间复杂度比较：
选择排序（蛮力法）：O（n2）
冒泡排序（蛮力法）：O（n2）
（改进的冒泡法其最坏情况是O（n2），最好情况是O（n））
归并排序：O（n log2n）
快速排序：O（n log2n）
插入排序：O（n2）
堆排序：O（n log2n）
若对于一个几乎排好序的序列，插入排序的最好情况为O（n），而快速排序为O（n log2n），故C对；
(D) Suppose T(n) is the runtime of resolving a problem with n elements, T(n) = Θ(1) if n = 1; T(n) = 2T(n/2) + Θ(n) if > 1; so T(n) is Θ(n log n).
(E) None of the above.
（BD（AC？））9. Which of the following statements are true?
(A) Insertion sort and bubble sort are not effcient for large data sets.
(B) Quick sort makes O(n^2) comparisons in the worst case.
快速排序最差为：O（n2）
(C) There is an array: 7, 6, 5, 4, 3, 2, 1. If using selection sort (ascending), the number of swap operation is 6.
(D) Heap sort uses two heap operations: insertion androot deletion.
堆排序用到两种堆操作：插入和根移除（√？）
(E) None of above.
（A？）10. Assume both x and y are integers, which one of the followings returns the minimum of the two integers?
^ 按位异或
如果x < y：A为0；如果x > y : A为(x ^ y)
B、C、D为(x ^ y)
(A) y ^ ((x ^ y) & ~(x < y))
(B) y ^(x ^ y)
(C) x ^ (x ^ y)
(D) (x ^ y) ^ (y ^ x)
(E) None of the above
（BCD）11. The Orchid Pavilion (兰亭集序) is well known as the top of "行书" in history of Chinese literature. The most fascinating sentence "Well I know it is a lie to say that life and death is the same thing, and that longevity and early death make no difference! Alas!" ("周知一死生为虚诞， 齐彭殇为妄作。") By counting the characters of the whole content (in Chinese version), the result should be 391 (including punctuation). For these characters written to a text file, please select the possible file size without any data corrupt.
兰亭集序有361个中文字符（含标点），请问存储在文本文件中的时候，文件大小可能是多大？
(A) 722字节 UTF-16 （这个不对，因为UTF-16有Big Endian和Little Endian两种，必须要加BOM）
(B) 724字节 UTF-16 （这个是对的，UTF-16两字节表示一个汉字，外加一个BOM两字节）
(C) 1083字节 UTF-8 （这个是对的，UTF-8通常三字节一个汉字，选用不加BOM的方式）
(D) 1086字节 UTF-8 （这个是对的，UTF-8通常三字节一个汉字，选用加BOM的方式）
(A) 782 bytes in UTF-16 encoding
(B) 784 bytes in UTF-16 encoding
(C) 1173 bytes in UTF-8 encoding
(D) 1176 bytes in UTF-8 encoding
(E) None of the above
（BC）12. Fill the blanks inside class definition
class Test
{
public:
____ int a;
____ int b;
public:
Test::Test(int _a, int _b) : a(_a) {b = _b;}
};
int Test::b;
int _tmain(int argc, __TCHAR *argv[])
{
Test t1(0, 0), t2(1, 1);
t1.b = 10;
t2.b = 20;
printf("%u %u %u %u", t1.a, t1.b, t2.a, t2.b);
}
Running result: 0 20 1 20
b 肯定是静态的
(A) static/const
(B) const/static
(C) --/static
(D) const static/static
(E) None of the above
（A）13. A 3-order B-tree has 2047 key words, what is the maximum height of the tree?
M阶B树只能在叶子结点存储数据，其他结点的孩子个数必须在[ceiling(M/2), M]之间，根节点要么是叶子结点，要么至少有两个孩子。
根据该定义，如果3阶B树有2048个元素，那高度最大时每个结点都取孩子个数下限(2)，高度为12（即log(2048)+1)。
然后去掉一个叶子结点，则从叶子向根一路发生结点合并，一个2孩子结点和一个1孩子结点合并成为一个3孩子结点。
这不是高潮，高潮是，根节点的两个孩子也发生了结点合并，变成了一个结点，根不再满足B树的要求，被删除，其唯一的孩子成为了新的根。
于是，树的高度变为了11
(A) 11 (B) 12 (C) 13 (D) 14
（ACE？）14. In C++, which of the following keyword(s) can be used on both a variable and a function?
在c++中,下列哪个关键字(s)既可用于表示变量，又可表示一个函数?
(A) static (B) virtual (C) extern (D) inline (E) const
（D）15. What is the result of the following program?
char* f(char *str, char ch)
{
char *it1 = str;
char *it2 = str;
while (*it2 != '\0') {
while (*it2 == ch) {
it2++;
}
*it1++ = *it2++;
}
return str;
}
void main(int argc, char *argv[])
{
char *a = new char[10];
strcpy(a, "abcdcccd");
cout << f(a, 'c');
}
(A) abdcccd (B) abdd (C) abcc (D) abddcccd (E) Access Violation
（A）16.Consider the following definition of a recursive function, power, that will perform exponentiation.
考虑以下定义的递归函数, power，将执行求幂；
int power(int b, int e)
{
if (e == 0) return 1;
if (e %2 == 0) return power (b * b, e / 2);
return b * power(b * b, e / 2);
}
Asymptotically (渐进地) in terms of the exponent e, the number of calls to power that occur as a result of the call power(b, e) is
(A) logarithmic (B) linear (C) quadratic (D) exponential
A、对数的
（B）17. Assume a full deck of cards has 52 cards, 2 black suits (spade and club) and 2 red suits (diamond and heart).
If you are given a full deck, and a half deck (with 1 red suit and 1 black suit), what's the possibility for each one getting 2 red cards if taking 2 cards?
(A) 1/2, 1/2 (B) 25/102, 12/50 (C) 50/51, 24/25 (D) 25/51, 12/25 (E) 25/51, 1/2
（D）18. There is a stack and a sequence of n numbers (i.e., 1, 2, 3, ..., n). Push the n numbers into the stack following the sequence and pop out randomly. How many different sequences of the number we may get? Suppose n is 2, the output sequence may be 1, 2 or 2, 1, so we get 2 different sequences.
(A) C_2n^n
(B) C_2n^n - C_2n^(n + 1)
(C) ((2n)!) / (n + 1)n!n!
(D) n!
(E) None of the above
（C）19. Longest Increasing Subsequence (LIS) means a sequence containing some elements in another sequence by the same order, and the values of elements keeps increasing.
For example, LIS of {2, 1, 4, 2, 3, 7, 4, 6} is {1, 2, 3, 4, 6}, and its LIS length is 5.
Considering an array with N elements, what is the lowest time and space complexity to get the length of LIS?
(A) Time: N^2, Space: N^2
(B) Time: N^2, Space: N
(C) Time: NlogN, Space: N
(D) Time: N, Space: N
(E) Time: N, Space: C
（E）20. What is the output of the following piece of C++ code?
#include
using namespace std;
struct Item
{
char c;
Item *next;
};
Item *Routine1(Item *x)
{
Item *prev = NULL, *curr = x;
while (curr) {
Item *next = curr->next;
curr->next = prev;
prev = curr;
curr = next;
}
return prev;
}
void Routine2(Item *x)
{
Item *curr = x;
while (curr) {
cout << curr->c << " ";
curr = curr->next;
}
}
void _tmain(void)
{
Item *x,
d = {'d', NULL},
c = {'c', &d},
b = {'b', &c},
a = {'a', &b};
x = Routine1(&a);
Routine2(x);
}
(A) cbad (B) badc (C) dbca (D) abcd (E) dcba