poj 3286 统计0的个数

 1 #include <iostream>
 2 
 3 using namespace std;
 4 long long p;
 5 long long  a[20];
 6 long long  solve(long long  n){
 7     long long left,m,sum =0;
 8     for(int i=1;i<12;i++){
 9         left = n/a[i]-1;
10         sum += left*a[i-1];
11         m = (n%a[i]-n%a[i-1])/a[i-1];
12         if(m>0) sum += a[i-1];
13         else if(m==0) sum += n%a[i-1]+1;
14         if(n<a[i]) break;
15     }
16     return sum;
17 }
18 
19 int main()
20 {
21     a[0] =1;
22     for(int i=1;i<15;i++)
23         a[i] =a[i-1]*10;
24     long long  m,n;
25     while(cin>>m>>n){
26         if(m<0)
27             break;
28         if(n==0){
29             cout<<1<<endl;
30             continue;
31         }
32         long long  res1 = solve(n);
33         long long res2 = solve(m-1);
34         //cout<<p<<endl;
35         //cout<<res1<<" "<<res2<<endl;
36         cout<<res1-res2<<endl;
37     }
38     return 0;
39 }
40 /**
41 if(n<0)
42         return 0;
43     long long  low;
44     long long  high;
45     long long  cur;
46     p =1;
47     int tmp=0;
48     long long cnt =0;
49     while(n/p!=0){
50         tmp ++;
51         low = n-(n/p)*p;
52         high = n/(p*10)-1;
53         cur = (n/p)%10;
54         switch (cur){
55         case 0:
56             cnt += high*p+low+1;
57             break;
58         default:
59             cnt += (high+1)*p;
60             break;
61         }
62         p =p*10;
63     }
64     //for(int i=0;i<tmp;i++){
65     //    cnt -= a[i];
66     //}
67     return cnt;
68 */