1.輸入n，打印一到n位最大數字，比如輸入3，打印1到999

面試題是我找的，代碼也看了，但是代碼我自己寫的！！！

//大數相加，相減 相除 相乘
#include <iostream>
using namespace std;

//此轉換只是在特定條件下的 int類型只有0-9 
inline int num(char s){return s-'0';}//char轉換成int
inline char ch(int num){return num+'0';}//int轉換成char

char * add(char a[],char b[]);

char c[100] ={'0'};
void main()
{
	char a[100] ={'0'} ; //保存第一個數
    char b[100] = {'0'}; //保存第二個數

	while(1){
	memset(a,'\0',100);
	memset(b,'\0',100);
	//i=1，j=1,保留符號位+和-
    int i=1,j=1;
	cout<<" 根據提示輸入整數，輸入非數字程序將會中止退出！ "<<endl;
    cout<<"請輸入第一個數字：";
	//獲取符號位
	a[0]=getchar();
	if(a[0] > '9'&&a[0] < '0'&&a[0] != '-'&&a[0] != '+')continue;
	else if(a[0] == '-')a[0] ='-';
	else if ( a[0]>='0' && a[0]<='9')
	{a[1] =a[0];a[0] = '+';i++;}
    while ( (a[i]=getchar()) !='\n' )
    {
          if ( a[i]>'9' && a[i]<'0')break;
          else{
			  i++;
          }
    }
	a[i]='\0';
	cout<<"請輸入第二個數字：";
	//獲取符號位
	b[0]=getchar();
	if(b[0] > '9'&&b[0] < '0'&&b[0] != '-'&&b[0] != '+')continue;
	else if(b[0] == '-')b[0] ='-';
	else  if ( b[0]>='0' && b[0]<='9')
	{b[1] =b[0];b[0] = '+';j++;}
    while ( (b[j]=getchar()) !='\n' )
    {
          if ( b[j]>'9' && b[j]<'0')break;
          else{
			  j++;
          }
    }
	b[j]='\0';
	cout<<a;
	cout<<b<<endl;
	cout<<add(a,b)<<endl;
	}

	system("pause");
}

char * add(char a[],char b[])
{
	int l =strlen(a)-1;
	int r =strlen(b)-1;
	int result;
	int overflow =0;
	if(l>=r)
	{
		for(int i=l;i>0;i--)
		{
			if(i>l-r){
				result =(num(a[i])+num(b[i-l+r])+overflow)%10;
				overflow =(num(a[i])+num(b[i-l+r])+overflow)/10;
				a[i] = ch(result);
			}
			else{
				result =(num(a[i])+overflow)%10;
				overflow =(num(a[i])+overflow)/10;
				a[i] = ch(result);
				if(i==1&&overflow!=0) 
				{
					memset(c,'\0',100);
					c[0] =a[0];
					c[1] =ch(overflow);
					for(int s =1;s<=l;s++)
						c[s+1]=a[s];
					return c;
				}
				
			}
		}
		return a;
	}else
	{
		for(int i=r;i>0;i--)
		{
			if(i>r-l){
				result =(num(b[i])+num(a[i-r+l])+overflow)%10;
				overflow =(num(b[i])+num(a[i-r+l])+overflow)/10;
				b[i] = ch(result);
			}
			else{
				result =(num(b[i])+overflow)%10;
				overflow =(num(b[i])+overflow)/10;
				b[i] = ch(result);
				if(i==1&&overflow!=0) 
				{
					memset(c,'\0',100);
					c[0] =b[0];
					c[1] =ch(overflow);
					for(int s =1;s<=l;s++)
						c[s+1]=b[s];
					return c;
				}
			}
		}
		return b;
	}
}