C编程小题

1：数组越界和unsigned char

#define MAX 255
void main()
{
unsigned char i;
  unsigned char A[MAX];
  for(i = 0; i <= MAX; i++)//i达到255后，加1变为0，无线循环下去
  {
    A[i] = i;
    printf("*%d*",A[i]);
  }
}

两个问题：数字越界和无限循环，char的范围[-128,127],unsigned char [0,255]，

--------------------------------------------------------------------------------------------------------------分割线----------------------------------------------------------------------------------------------------------------

2：程序

void GetMemory(char *p)
{
    p = (char *)malloc(100);
}

void Test(void)
{
    char *str = NULL;
    GetMemory(&str);
    strcpy(str,"hello world");
    printf("%s",str);
}

void GetMemory(char **p,int num)
{
    *p = (char *)malloc(num);
}

void Test(void)
{
    char *str = NULL;
    GetMemory(&str,100);
    strcpy(str,"hello world");
    printf("%s",str);
}

注意这些对内存的考察主要集中在，指针的理解和变量的生存期及作用范围，良好的动态内存申请，释放习惯，free，delete后置空是最常见的操作

--------------------------------------------------------------------------------------------------------------分割线----------------------------------------------------------------------------------------------------------------

3：指针问题

swap()
{
   int *p;//p是一个野指针，应该为int p
   *p = *p1;
    *p1=*p2;
    *p2 = *p;
}

--------------------------------------------------------------------------------------------------------------分割线----------------------------------------------------------------------------------------------------------------

4:冒泡排序

void BubleSort(int p[],int n)
{
    int i = 0;
    int j = 0;
    int change_flag = 0;
    int temp = 0;
    for(j = n; j > 0;j--)
    {
        for(i = 0; i < j; i++)
        {
            if(p[i] > p[i+1])
            {
                temp = p[i];
                p[i] = p[i+1];
                p[i+1] = temp;  
                change_flag = 1;
            }  
        }
        if(change_flag == 0)break;
    }
}

--------------------------------------------------------------------------------------------------------------分割线----------------------------------------------------------------------------------------------------------------

5:进制转换

十进制数N和其他进制数的转换，其解决方法有很多，其中一个简单算法基于下列原理： N = （N div D）*D  + N mod D

例如：十进制1348 转8进制等于2504，

N          N div 8      N mod 8

1348      168             4

16821    0

212    5

20    2

注意有个倒置的过程可以利用栈结构

void conversion()
{
	InitStack(S);
	while (N) {
		Push(S,N%8);
		N = N / 8;
	}
	while (!StackEmpty(S)) {
		Pop(S,e);
		printf("%d",e);
	}
}

m进制转为n进制

void m2n(int m, char* mNum, int n, char* nNum)   
{  
    int i = 0;  
    char c, *p = nNum;  
  
    //这是一个考察地方，是否能用最少乘法次数。  
    while (*mNum != '\0')  
        i = i*m + *mNum++ - '0';  
      
    //辗转取余  
    while (i) {  
        *p++ = i % n + '0';  
        i /= n;  
    }  
    *p-- = '\0';  
  
    //逆置余数序列  
    while (p > nNum) {  
        c = *p;  
        *p-- = *nNum;  
        *nNum++ = c;  
    }  
}  

--------------------------------------------------------------------------------------------------------------分割线---------------------------------------------------------------------------------------------------------------

6.求阶 阶乘问题的递归算法

long int fact(int n)
{
	if(n == 0|| n == 1)
		return 1;
	else
		return n*fact(n - 1);
}

7c++小题

#include <iostream>
using namespace std;
class human
{
public:
	human(){
		cout<<"构造"<<endl;
		human_num++;}; //默认构造函数
	static int human_num;//静态数据成员
	~human()
	{
		cout<<"析构函数"<<endl;
		human_num--;
		print();
	}
	void print()
	{
		cout<<"human num is:"<<human_num<<endl;
	}
protected:
private:
};
int human::human_num = 0;//类中静态数据成员在外部定义，仅定义一次
human f1(human x)
{
	x.print();
	return x;
}
int main()
{
	human h1;
	h1.print();
	cout<<"-----1----"<<endl;
	human h2 = f1(h1);//先调用f1(),输出human_num：1，而后输出human_num 为0，f1是按值传递对象，调用默认的复制构造函数，h2没有调用定义的构造函数。
	cout<<"-----2----"<<endl;
	h2.print();
	system("pause");
	return 0 ;
}

结果：

构造
human num is:1
-----1----
human num is:1
析构函数 //return 调用析构函数
human num is:0
-----2----
human num is:0
请按任意键继续. .

--------------------------------------------------------------------------------------------------------------分割线---------------------------------------------------------------------------------------------------------------

8：大数相乘

#include <iostream>
using namespace std;
#define N 30
int a[N],b[N],c[2*N];
void strToNum(int *p,string &s)//字符串放入数组
{
	for (int i = 0; i <=(s.length() - 1);i++)
	{
		p[i] = s[s.length() - 1 - i] - '0';
	}
}
void multiplay(string&s1,string &s2)//相乘
{
	strToNum(a,s1);
	strToNum(b,s2);
	for (int i = 0; i < N; i++)cout<<a[i];
	cout<<endl;
	for (int i = 0; i < N; i++)cout<<b[i];
	cout<<endl;
	for (int i = 0; i < 2*N; i++ )
	{
		c[i] = 0;
	}
	for (int i = 0; i < N; i++)
		for (int j = 0; j < N; j++)c[i+j] += a[i]*b[j]; //这句是大数相乘的关键
	for (int i = 0; i < 2*N-1; i++)//处理进位
	{	
		int m = c[i]/10;
		c[i] = c[i]%10;
		c[i+1] += m;
	}
}
int main()
{
	string s = "7898912";
	string s1 = "12312";
	multiplay(s,s1);
	for (int i = 2*N - 1; i >= 0;i--)
	{
		cout<<c[i];
	}
	system("pause");
	return 0;
}:

9：子数组最大值问题

#include <iostream>
using namespace std;

int a[10] = {1,3,-4,5,9,-10,3,4,1,-5};
int summax(int *a,int len)
{
	int nStart = a[len - 1];
	int nAll = a[len - 1];
	for (int i = len - 2;i>=0; i--)
	{
		if (nStart < 0)nStart = 0;
		nStart += a[i];
		if (nStart > nAll)nAll = nStart;
	}
	return nStart;
}

int main()
{
	int m = summax(a,10);
	cout<<m<<endl;
	system("pause");
	return 0;
}