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]，

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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

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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;
}

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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;
}
}

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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;
}
}

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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. .

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8：大数相乘

#include <iostream>
using namespace std;
#define N 30
int a
,b
,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;
}