UVa 10706 - Number Sequence

10706 - Number Sequence

 * http://acm.uva.es/p/v107/10706.html

Summary
This problem involves a little bit of number theory and can be solved using binary search.

Explanation
Let the sequence $$A_k$$ be $$S_1 S_2 ... S_k$$, where $$S_k$$ consists of positive integer numbers ranging from 1 to k. A natural question is, what is the length or the number of digits of $$A_k$$. It is not hard to write a small subroutine to compute the length of $$A_k$$. Let $$len(S)$$ denote the length of some sequence S and $$d_k$$ the number of digits of k. Then by noticing that $$ len(S_k) = len(S_{k-1}) + d_k $$ we have $$ len(A_k) - len(A_{k-1}) = len(S_k) = len(S_{k-1}) + d_k = len(S_{k-2}) + d_{k-1} + d_k = ... = len(S_1) + \sum_{i=2}^k d_k $$. Since $$d_k$$ is bounded in this problem, it can be estimated that $$ len(A_k) = O(k^2) $$. This rough estimation helps us to pinpoint the biggest possible k value involved in this problem. Actually $$ len(A_{31267}) < 2,147,483,647 < len(A_{31268}) $$. With the knowledge of the maximum k value, given the index i of a certain digit, we can use the binary search to locate the $$S_k$$ to which the digit belong. This technique is similar to the one used in quick sort and thus the details are ignored. After locating the $$S_k$$, the rest of the job is trivial.

Gotchas

 * Take care of the marginal cases properly.

Input
18 8 3 80 546 23423 65753 2345 45645756 546454 6786797 131231 78934124 68904565 123487907 5655 778888 101011 2147483647

Output
2 2 0 2 3 1 5 5 2 5 9 7 5 7 1 5 5 2