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    微软 2021 暑期实习面试题

    小编:管理员 1012阅读 2021.10.11

    第1题:

    Selection sort 80 items, after 32 iterations, how many positions of items are determined?



    第2题:

     Which is used in sync process//thread in operation system?
    A. Mutex             B.mailbox        C.Semaphore        D.local procedure call



    第3题:

     Size of a stack is 5, input is a sequence 1, 2, …, 7, which is possible output?
    A. 1234567       B. 7654321       C.5643721       D. 1765432       E. 3217564



    第4题:

    010111001 * 011001 + 1101110 = ?
    A. 0001010000111111       B. 0101011101110011       C. 0011010000110101 



    第5题:

    What is the output of the follow code?
    void main(int argc, char* argv[]) {
    int i = 11;
    int const *p = &i;
    p++;        
    cout<<*p<<endl;}
    A. 11        B. 12        C. Garbage value        D. Comipler error        E. None of above



    第6题:

    Which number has difference 78633 after 180-degree rotation?
    A. 60918                B.91086                C. 18609                D. 10968                E.86901



    第7题:

    Which statement is true?
    A. Inorder and preorder can determine a Binary tree
    B. Postorder and preorder can determine a Binary tree
    C. For almost sorted array, Insertion sort is more efficient than Quicksort
    D. If T(n)=2T(n/2)+?(n), Then T(n)=?(nLogn)
    E. none of above



    第8题:

    Which statement is true?
    A. Insertion and buble sort are not effient for large data sets
    B. The complexity of Quick Sort is O(n2) in worst case
    C. It is needed 6 swap operations to sort sequence 7,6,5,4,3,2,1(ascending) by Selection sort
    D. Heap sort has two operations: Insertion and root deletion
    E. None of above



    第9题:

    The output of the following code is 0 20 1 20, what are the type of a and b?
    class Test{
    ____ int a;
    ____ int b;
    Test(int _a, int _b) {a = _a; b=_b;}
    }
    void main() {
    Test t1(0, 0), t2(1,1);
    t1.b = 10;
    t2.b = 20;
    cout<<t1.a<<” ” <<t1.b<<” ” <<t2.a<<” ” <<t2.B<<” ”<<endl;
    }
    A. static/const        B. const/static        C. __/static        D. const static/static        E. None of
    above



    第10题:

     A 3-order B-tree has 2047 key words, what is the maximum height?
    A. 11          B. 12          C. 13          D. 14



    第11题:

     Which can be used both to variable and function?
    A. static        B. virtual                C. extern        D. inline        E. const



    第12题:

     What is the output of the follow code?
    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(int argc, char* argv[]) {
    char *a = new char[10];
    strcpy(a, "abcdcccd");
    cout<<f(a, 'c');
    return 0;
    }

    A. abdcccd        B. abdd        C. abcc        D. abddcccd        E. Access violation



    第13题:

    What is the complexity of the result call of power(b, e) in the follow code?
    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);
    }
    A. logarithmic                B. linear        C. quadratic        D. exponentical



    第14题:

    Take 2 cards from one full poker(52 cards, 26 red and 26 black) and half poker each, what is the probability of the event that two cards are both red?
    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



    第15题:

    How many kinds of output of stack with the input 1,2,…,n?
    B. C_2n^n-C_2n^(n+1)                C. ((2n)!)/(n+1)n!n!        D. n!        E. none



    第16题:

    What is the minimum time and space complexity to compute the Largest Increased Subsequence(LIS) of array?
    A. N^2,N^2          B. N^2,N                  C. NlogN,N            D. N, N        E. N,C



    第17题:

    What is the output of the follow code?
    struct Item{
    char c;
    Item *next;
    };
    Item* f1(Item* x){
    Item *prev = NULL;
    Item *curr = x;        
    while(curr) {                Item *next = curr->next;                
    curr->next = prev;                
    prev = curr;
                    curr = next;
    }        
    return prev;
    }
    void f2(Item *x){
    while(x){
    cout<<x->c;
    x = x->next;        
    }
    }
    int main(int argc, char* argv[]){
    Item *x, d = {'d', NULL}, c = {'c', &d}, b = {'b', &c}, a = {'a', &b};        
    x = f1(&a);
    f2(x);        
    return 0;
    }


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