UVa 11038 - How Many O's?

Summary
Write down all integers between $$m$$ and $$n$$ inclusive ($$m \le n$$) in decimal. How many 0's will you write?

Explanation
Notation: $$\overline{a_n a_{n-1} \dots a_0}$$ denotes an integer $$a = a_n 10^n + a_{n-1} 10^{n-1} + \cdots + a_0$$. That is $$a_n,\ a_{n-1},\ \dots,\ a_0$$ are the decimal digits of $$a$$, from left to right.

Let's solve an easier problem. How many 0's are there in numbers between 0 and $$b$$ inclusive? If we denote this number by $$f(b)$$ then the answer to our original problem is just $$f(n) - f(m-1)$$.

Let $$b = \overline{b_n b_{n-1} \dots b_0}$$. Let's find for each position $$0 \le k < n$$ how many times a zero appears there as we are counting from 0 to $$b$$.

If $$b_k > 0$$, then by setting $$a_k = 0$$, and choosing the other digits according to the constraints: $$1 \le \overline{a_n \dots a_{k+1}} \le \overline{b_n \dots b_{k+1}}$$, and $$0 \le \overline{a_{k-1} \dots a_0} < 10^k$$, we will have a positive integer $$a = \overline{a_n \dots a_0}$$, which is not greater than $$b$$, and the $$k$$-th digit of which exists and is equal to zero. There are $$10^k \cdot \overline{b_n \dots b_{k+1}}$$ such integers.

If $$b_k = 0$$, same analysis as above applies, except that when $$\overline{a_n \dots a_{k+1}} = \overline{b_n \dots b_{k+1}}$$ there are only $$1 + \overline{b_{k-1} \dots b_0}$$ ways to choose the digits to the right of $$a_k$$. So in this case there are $$(10^k) \cdot (\overline{b_n \dots b_{k+1}} - 1) + 1 + \overline{b_{k-1} \dots b_0}$$ integers between 0 and $$b$$, in which the $$k$$-th digit is zero.

The total number of zeroes is the sum of the number of times a zero occurs in each position, plus 1 for the integer "0".

Input
10 11 100 200 0 500 1234567890 2345678901 0 4294967295 1001010101 1010101010 3020203 302010203 -1 -1

Output
1 22 92 987654304 3825876150 23709830 231396207

Implementations

 * Sweepline - C++