1到N中1出现的个数

#include <iostream>
using namespace std;

int count_one(int n)
{
if(n<1)
return 0;
int round = n;
int base = 1;
int count = 0;
while(round>0)
    {
int weight = round%10;
       round/=10;
//count+=round;
if(weight==1)
          count+=round*base+(n%10)+1;
else if(weight>1)
          count+=round*base+base;
else if(weight==0)
            count+=round*base;
base*=10;
    }
return count;
}


/*
 若weight为0，则1出现次数为round*base
 若weight为1，则1出现次数为round*base+former+1
 若weight大于1，则1出现次数为rount*base+base
*/
int main()
{
int n;
while(cin>>n)
    {
        cout<<count_one(n)<<endl;
    }
return 0;
}