找出十进制数中出现的''一''的个数

 1 #include "iostream.h"
 2 #include "stdlib.h"
 3 int CountNum(int n)
 4 {
 5     int count=0; 
 6     int factor=1; 
 7     int LowerNum=0;
 8     int CurNum=0;
 9     int HigherNum=0;
10     while (n/factor!=0)       
11     {
12         LowerNum=n-(n/factor)*factor;
13         CurNum=(n/factor)%10;
14         HigherNum=n/(factor*10);
15         switch (CurNum)
16         {
17         case 0:
18             count=count+HigherNum*factor;
19             break;
20         case 1:
21             count=count+HigherNum*factor + LowerNum +1;
22             break;
23         default:
24             count=count+(HigherNum+1)*factor;
25             break;
26         }
27         factor=factor*10;
28     }
29     return count;
30 }
31 
32 int main()
33 {
34     int num;
35     cout<<"请输入数字:\n";
36     cin>>num;
37     cout<<"\n";
38     cout<<num<<"中出现数字1的个数为：\n";
39     cout<<CountNum(num)<<endl;
40     return 0;
41 }