微軟校園招聘筆試題
阿新 • • 發佈：2018-12-20
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 C、32 D、39 E、402、Which synchronization mechanism(s) is/are used to avoid race conditions among processes/threads in operating system?A、Mutex 
C、0011010000110101
轉化為10進位制操作以後，再轉化為二進位制就可以了。
5、What is output if you compile and execute the following c code?
void main(){ int i = 11; int const 6、Which of following C++ code is correct ? C
A、
int f(){ int *a = new int(3); return *a;}int *f(){ int a[3] = {1,2,3}; return a;}vector<int> f(){ vector<int> v(3); return v;}void f(int *ret){ int a[3] = {1,2,3}; ret = a; return ;}E、None of above
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？A、60918 B、91086 C、18609 D、10968 E、86901
8、Which of the following statements are trueA、We can create a binary tree from given inorder and preorder traversal sequences.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 Quciksort.D、Suppose T(n) is the runtime of resolving a problem with n elements, T(n)=O(1) if n=1;T(n)=2*T(n/2)+O(n) if n>1; so T(n) is O(nlgn)E、None of above9、Which of the following statements are true?A、Insertion sort and bubble sort are not efficient for large data sets.B、Qucik sort makes O(n^2) comparisons in the worst case.C、There is an array :7,6,5,4,3,2,1. If using selection sort (ascending),the number of swap operations is 6D、Heap sort uses two heap operations:insertion and root deletion （插入、堆調整）E、None of above10、Assume both x and y are integers,which one of the followings returns the minimum of the two integers?A、y^((x^y) & -(x<y))B、y^(x^y)C、x^(x^y)D、(x^y)^(y^x)E、None of above
x<y的時候，-（x<y）即-1的補碼形式就是全1（111111），（x^y）&-（x<y）== x^y。x>y的時候，-（x<y）即0的補碼形式就是全0（000000），（x^y）&-（x<y）== 0
11、The Orchid Pavilion(蘭亭集序) is well known as the top of “行書”in history of Chinese literature. The most fascinating sentence is "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!"(固知一死生為虛誕，齊彭殤為妄作).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.A、782 bytes in UTF-16 encodingB、784 bytes in UTF-16 encodingC、1173 bytes in UTF-8 encodingD、1176 bytes in UTF-8 encoding
E、None of above
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 main(void){ 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); return 0;} A、static/const
B、const/static
C、--/static
D、conststatic/static
E、None of above
13、A 3-order B-tree has 2047 key words,what is the maximum height of the tree?A、11 B、12 C、13 D、14解析：m階B-樹的根節點至少有兩棵子樹，其他除根之外的所有非終端節點至少含有m/2(向上取整)棵子樹，即至少含有m/2-1個關鍵字。根據題意，3階的B-樹若想要達到最大的高度，那麼每個節點含有一個關鍵字，即每個節點含有2棵子樹，也就是所謂的完全二叉樹了，這樣達到的高度是最大的。即含有2047個關鍵字的完全二叉樹的高度是多少，這也是為什麼這種題只出3階的原因吧，就是為了轉化成求完全二叉樹的深度。很明顯求得高度是11，但是由於B-樹還有一層所謂的葉子節點，可以看作是外部結點或查詢失敗的結點，實際上這些結點不存在的，指向這些結點的指標為空。所以不考慮葉子節點資訊的時候，最大高度是11，考慮葉子節點資訊的時候，最大高度就是12了。14、In C++,which of the following keyword(s）can be used on both a variable and a function?A、static B、virtual C、extern D、inline E、const15、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;}int main(void){ char *a = new char[10]; strcpy(a , "abcdcccd"); cout<<f(a,'c'); return 0;}16、Consider the following definition of a recursive function,power,that will perform exponentiation.
int power(int b , int e){ if(e == 0) return 1; if(e % 2 == 0) return power(b*b , e/2); else return b * power(b*b , e/2);}20、What is the output of the following piece of C++ code ？
#include<iostream>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; }}int main(void){ Item *x, d = {'d' , NULL}, c = {'c' , &d}, b = {'b' , &c}, a = {'a' , &b}; x = Routine1( &a ); Routine2( x ); return 0;}#include<iostream>using namespace std;class Base{public: char Value() { return 'A';} virtual char VirtualValue() { return 'X';}};class Derived:public Base{public: char Value(){return'U';}};class VirtualDerived:virtualpublic Base{public: char Value() { return 'Z';} char VirtualValue() { return 'V';}};void main(){ Base *p1=new Derived(); Base *p2=new VirtualDerived(); cout<<p1->Value()<<" "<< p1->VirtualValue()<<" "<< p2->Value()<<" "<< p2->VirtualValue()<<endl;}