poj 3286 统计0的个数

 #include <iostream>

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

 int main()
 {
     a[] =;
     for(int i=;i<;i++)
         a[i] =a[i-]*;
     long long  m,n;
     while(cin>>m>>n){
         if(m<)
             break;
         if(n==){
             cout<<<<endl;
             continue;
         }
         long long  res1 = solve(n);
         long long res2 = solve(m-);
         //cout<<p<<endl;
         //cout<<res1<<" "<<res2<<endl;
         cout<<res1-res2<<endl;
     }
     return ;
 }
 /**
 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;
 */