pat 1049. Counting Ones (30)

#include<iostream>
#include<stdio.h>
using namespace std;
int getone(int n)
{
int ans=0,base=1,right,left,now;
while(n/base)
{
right=n%base;
left=n/(base*10);
now=(n/base)%10;
if(now==0)ans+=left*base;
else
if(now==1)ans+=left*base+right+1;
else
ans+=(left+1)*base;
base*=10;
}
return ans;
}
int  main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
printf("%d\n",getone(n));
}
return 0;
}