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 */

poj 3286 统计0的个数