Poj 3286 How many 0's?

 1 #include<stdio.h>
 2 #include<string.h>
 3 using namespace std;
 4 typedef long long LL;
 5 LL ab[20];
 6            
 7 LL Solve(LL n)
 8 {
 9     LL sum=1;
10     if(n<0) return 0;
11     for(int i=1;i;i++)
12     {
13         if(ab[i]>n) break;
14         LL front=n/ab[i];
15         LL after=n%ab[i-1];
16         LL now=(n%ab[i]-n%ab[i-1])/ab[i-1];
17         if(now==0) sum+=(front-1)*ab[i-1]+after+1;
18         else sum+=front*ab[i-1];
19     }
20     return sum;
21 }
22            
23 int main()
24 {
25     LL a,b;
26     ab[0]=1;
27     for(int i=1;i<15;i++) ab[i]=ab[i-1]*10;
28     while(scanf("%lld%lld",&a,&b)!=EOF)
29     {
30         if(a==-1 && b==-1) break;
31         LL ans1=Solve(a-1);
32         LL ans2=Solve(b);
33         printf("%lld\n",ans2-ans1);
34     }
35     return 0;
36 }