POJ 3286 How many 0's?

A Benedict monk No.16 writes down the decimal representations of all natural numbers between and including m and n, m ≤ n. How many 0‘s will he write down?

Input consists of a sequence of lines. Each line contains two unsigned 32-bit integers m and n, m ≤ n. The last line of input has the value of m negative and this line should not be processed.

For each line of input print one line of output with one integer number giving the number of 0‘s written down by the monk.

#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdio>
#define N 30
using namespace std;
__int64 dp1[N],dp2[N];
int dig[N];
int Top;
__int64 ba[N];
int main()
{
    //freopen("data.txt","r",stdin);
    __int64 f();
    memset(dp1,0,sizeof(dp1));
    memset(dp2,0,sizeof(dp2));
    dp1[1] = dp2[1] = 1;
    ba[0] = 1;
    ba[1] = 10;
    for(int i=2;i<=11;i++)
    {
        dp1[i] = dp2[i-1]*9;
        dp2[i] = 10*dp2[i-1]+ba[i-1];
        ba[i] = ba[i-1] * 10;
    }
    __int64 n,m;
    while(scanf("%I64d %I64d",&n,&m)!=EOF)
    {
        if(n<0)
        {
            break;
        }
        Top = 0;
        n-=1;
        __int64 res1,res2;
        if(n==-1)
        {
            res1 = 0;
        }else
        {
            if(n==0)
            {
                dig[Top++] = 0;
            }
            while(n!=0)
            {
               dig[Top++] = n%10;
               n = n/10;
            }
            res1=f();
        }
        Top = 0;
        if(m==0)
        {
            dig[Top++] = 0;
        }
        while(m!=0)
        {
            dig[Top++] = m%10;
            m = m/10;
        }
        res2 = f();
        printf("%I64d\n",res2-res1);
    }
    return 0;
}
__int64 f()
{
    __int64 res = 0;
    for(int i=1;i<=Top-1;i++)
    {
        res+=dp1[i];
    }
    //cout<<res<<endl;
    int t;
    for(int i=Top-1;i>=0;i--)
    {
        if(i==Top-1&&Top>1)
        {
            t = 1;
        }else
        {
            t = 0;
        }
        for(int j=t;j<=dig[i]-1;j++)
        {
            res+=dp2[i];
            if(j==0)
            {
                res+=ba[i];
            }
        }
    }
    for(int i=Top-1;i>=0;i--)
    {
        if(dig[i]==0)
        {
            __int64 sum = 0;
            for(int j=i-1;j>=0;j--)
            {
                sum = sum*10+dig[j];
            }
            sum+=1;
            res+=sum;
        }
    }
    return res;
}