poj 3286 统计0的个数

#include <iostream>

using namespace std;
long long p;
long long  a[20];
long long  solve(long long  n){
    long long left,m,sum =0;
    for(int i=1;i<12;i++){
        left = n/a[i]-1;
        sum += left*a[i-1];
        m = (n%a[i]-n%a[i-1])/a[i-1];
        if(m>0) sum += a[i-1];
        else if(m==0) sum += n%a[i-1]+1;
        if(n<a[i]) break;
    }
    return sum;
}

int main()
{
    a[0] =1;
    for(int i=1;i<15;i++)
        a[i] =a[i-1]*10;
    long long  m,n;
    while(cin>>m>>n){
        if(m<0)
            break;
        if(n==0){
            cout<<1<<endl;
            continue;
        }
        long long  res1 = solve(n);
        long long res2 = solve(m-1);
        //cout<<p<<endl;
        //cout<<res1<<" "<<res2<<endl;
        cout<<res1-res2<<endl;
    }
    return 0;
}
/**
if(n<0)
        return 0;
    long long  low;
    long long  high;
    long long  cur;
    p =1;
    int tmp=0;
    long long cnt =0;
    while(n/p!=0){
        tmp ++;
        low = n-(n/p)*p;
        high = n/(p*10)-1;
        cur = (n/p)%10;
        switch (cur){
        case 0:
            cnt += high*p+low+1;
            break;
        default:
            cnt += (high+1)*p;
            break;
        }
        p =p*10;
    }
    //for(int i=0;i<tmp;i++){
    //    cnt -= a[i];
    //}
    return cnt;
*/