1 #include <iostream>
 2 #include <string>
 3 using namespace std;
 4 enum flag{
 5     valid=0,
 6     invalid=1
 7 };
 8 int judge=valid;
 9
10
11 long long paraInt(string str)
12 {
13     judge=valid;
14     bool Fuflag=false;
15     long long num=0;
16     int i=0;
17     if(str.size()==0)
18     {
19         judge=invalid;
20         return invalid;
21     }
22     if (str[0]=='-')
23     {
24         i=1;
25         Fuflag=true;
26         if(str.size()==1)
27         {
28             judge=invalid;
29             return invalid;
30         }
31     }
32
33     for (i;i<str.size();++i)
34     {
35         int temp=str[i]-'0';
36         if(temp>=0&&temp<=9)
37             num=num*10+temp;
38         else{
39             judge=invalid;
40             return invalid;
41         }
42     }
43     if(Fuflag)
44         a=-a;
45     return num;
46 }
47 int main()
48 {
49     string str;
50     while(cin>>str)
51     {
52         long long num=paraInt(str);
53         if(!judge)
54             cout<<num<<endl;
55         else
56         {
57             cout<<"invalid input"<<endl;
58         }
59         str.clear();
60     }
61     return 0;
62 }

//整形转成字符串函数实现
//题目不难，重点考察面试者对问题考虑的全面程度
#include <iostream>
using namespace std;
void itoa_mf(int num,char str[])
{  int sign = num;  int i = 0;  int j = 0;  char temp[100];  //如果是负数就去掉符号,将-1234转成1234  if(sign < 0)  {  num = -num;  }  //转成字符串，1234转成"4321"  do  {  temp[i] = num % 10 + '0';  num /= 10;  i++;  }while(num > 0);  //如果是负数的话，加个符号在末尾，如："4321-"  if(sign < 0)  {  temp[i++] = '-';  }  temp[i] = '\0';  i--;  //将temp数组中逆序输入到str数组中  //将"4321-" ====> "-1234"  while(i >= 0)  {  str[j] = temp[i];  j++;  i--;  }  //字符串结束标识  str[j] = '\0';
}  //测试用例
void main()
{  int a = 0;  char s[100];  itoa_mf(a,s);  cout << s << endl;  }