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c++一些筆試題目和整理的答案
NO1
Below is usual way we find one element in an array
const int *find1(const int* array, int n, int x)
{
const int* p = array;
for(int i = 0; i < n; i++)
{
if(*p == x)
{
return p;
}
++p;
}
return 0; }
In this case we have to bear the knowledge of value type "int", the size of array, even the existence of an array. Would you re-write it using template to eliminate all these dependencies?
template
const T *find1(const T* array, int n, T x)
{
const T* p = array;
for(int i = 0; i < n; i++)
{
if(*p == x)
{
return p;
}
++p;
}
return 0; }
NO2
Give an example of implementing a Stack in the template way(only template class declaration without detail definition and realization)
template
class Stack
{
public:
Stack(int = 10) ;
~Stack() { [] stackPtr ; }
int push(const T&);
int pop(T&) ;
int isEmpty()const { return top == -1 ; }
int isFull() const { return top == size - 1 ; }
private:
int size ; // number of elements on Stack.
int top ;
T* stackPtr ;
} ;
NO3
Implement the simplest singleton pattern(initialize if necessary).
class Singleton {
public:
static Singleton* Instance();
protected:
Singleton();
private:
static Singleton* _instance;
}
// Implementation
Singleton* Singleton::_instance = 0;
Singleton* Singleton::Instance() {
if (_instance == 0) {
_instance = new Singleton;
}
return _instance;
}
NO4
1.Jeff and Diamond like playing game of coins, One day they designed a new set of rules:
1)Totally 10 coins
2)One can take away 1,2or 4 coins at one time by turns
3)Who takes the last loses.
Given these rules Whether the winning status is pre-determined or not
解答： 1：從后面開(kāi)始考慮，最后肯定要留1個(gè)才能保證自己贏
2：所以要設(shè)法讓對(duì)方留下2，3，5個(gè)
3：也就是要自己取后留下1，4，6，7，8，9
4：如果自己取后留下6，對(duì)方取2個(gè)，與(3)矛盾，所以排除6
5：如果自己取后留下8，對(duì)方取4個(gè)，與(3)一樣情況，所以也排除8
6：同樣，9也不行，如果我抽后剩下9，對(duì)方抽2個(gè)，就反過(guò)來(lái)成對(duì)方抽剩成7個(gè)了，也與3)矛盾，所以也排除
7：所以很顯然，我只能抽剩1，4，7
8：因?yàn)橹荒艹楹笫?，4，7才能贏，我先抽得話不可能達(dá)到這幾個(gè)數(shù)，很顯然，只能讓對(duì)
方先抽，也即是先抽的人輸
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