sicily 1370 How many 0's? 递推 规律

//找规律 算m到n有数字中有几个0，包括m n
//先设数组 dp[i]表示 i长度，第一个固定且非0的所有数字的0的个数
//基于dp从，dp和可以直接算 0到1000000 ，不包括100000
//剩下的数字由高位向低位算，分0和非0去暴力就可以了
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;long long m, n;
long long result;
long long dp[17];long long act[17];
long long f( long long k )
{long long k2 = k;if( k == -1 ) return 0;long long result = 0;long long temp[34], m = 0;while( k ){temp[m++] = k % 10;k /= 10;}result += 1;for( int i = 2;i <= m-1;i++ )result += 9*dp[i];result += (temp[m-1]-1) * dp[m];for( int i = m-2;i >= 0;i-- ){if( temp[i] > 0 ) result += ( act[i] + (temp[i])*dp[i+1] );else              result += k2 % act[i]+1 ;}return result;
}
int main()
{act[0] = 1;for( int i = 1;i <= 16;i++ ) act[i] = act[i-1]*10;dp[1] = 0; dp[2] = 1;for( int l = 3;l <= 16;l++ ){dp[l] = dp[l-1]*10 + act[l-2];}while( scanf("%lld%lld",&m,&n ) && m >= 0 ){//f(n);printf( "%lld\n",f(n)-f(m-1) );}return 0;
}