课堂练习——找十进制数中数字1的个数

#include "stdafx.h"
#include<iostream.h>
#include "stdlib.h"
int CountNum(int n)
{
    int count=0; 
    int factor=1; 
    int LowerNum=0;
    int CurNum=0;
    int HigherNum=0;
    while (n/factor!=0)       
    {
        LowerNum=n-(n/factor)*factor;
        CurNum=(n/factor)%10;
        HigherNum=n/(factor*10);
        switch (CurNum)
        {
        case 0:
            count=count+HigherNum*factor;
            break;
        case 1:
            count=count+HigherNum*factor + LowerNum +1;
            break;
        default:
            count=count+(HigherNum+1)*factor;
            break;
        }
        factor=factor*10;
    }
    return count;
}

int main()
{
    int num;
    cout<<"请输入数字:\n";
    cin>>num;
    cout<<"\n";
    cout<<num<<"中出现数字1的个数为：\n";
    cout<<CountNum(num)<<endl;
    return 0;
}