概要：[cpp] view plaincopyprint?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;}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);
微软校园招聘笔试题,标签：笔试大全,http://www.88haoxue.com[cpp] view plaincopyprint?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;
}
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;
} Running result : 0 20 1 20
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阶B-树的根节点至少有两棵子树，其他除根之外的所有非终端节点至少含有m/2(向上取整)棵子树，即至少含有m/2-1个关键字。根据题意，3阶的B-树若想要达到最大的高度，那么每个节点含有一个关键字，即每个节点含有2棵子树，也就是所谓的完全二叉树了，这样达到的高度是最大的。即含有2047个关键字的完全二叉树的高度是多少，这也是为什么这种题只出3阶的原因吧，就是为了转化成求完全二叉树的深度。很明显求得高度是11，但是由于B-树还有一层所谓的叶子节点，可以看作是外部结点或查找失败的结点，实际上这些结点不存在的，指向这些结点的指针为空。所以不考虑叶子节点信息的时候，最大高度是11，考虑叶子节点信息的时候，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、const
15、What is the result of the following program?
[cpp] view plaincopyprint?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<
return 0;
}
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<
return 0;
}A、abdcccd
B、abdd
C、abcc
D、abddcccd
E、Access violation
16、Consider the following definition of a recursive function,power,that will perform exponentiation.
[cpp] view plaincopyprint?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);
}
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);
}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
17、Assume a full deck of cards has 52 cards,2 blacks 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 is 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
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 n numbers we may get? Suppose n is 2 , the output sequence may 1,2 or 2,1, so wo 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 above
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.
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