一种算法，查找给定限制“ l”（0 <= l <= 10 ^ 16）中的数字数量，其中至少出现一次数字“ n”？

Question

First of all, this is not a homework question. I was going through some combinatorial problems to solve algorithmically and I really just need to get a start towards solving this problem.

My initial thought was to declare an array of size 17 initialize it and run a loop to find the occurrence of a number, say '5' using a simple search. But the solution looks tedious and ugly.

1. Any thoughts on how to represent a big number (10^16)?
2. Is there a simple combinatorics formula/algorithm to solve this kind of problem?

Thanks.

Answer 1

Assuming you mean 0 <= l < 10^16 (not <= 10^16), all integers in this range have 10 digits (allowing for leading 0s). There are 10^16 total values in this range. I'd write the problem as:

Number with digit n = 10^16 - number WITHOUT digit n .

So how many ways can we not choose n in the 1's place? 9 ways. How many ways can we not put an 'n' in the 10s OR 1's place? 9*9. Following this logic, there are 9^16 ways of not putting an n in any of the 16 possible slots.

So your answer is 10^16 - 9^16.

If you actually meant 0 <= l <= 16, that range has only one more number, namely 10^16. The leading digit of this number is 1, so if n = 1 you have exactly 10^16 - 9^16 + 1 values with a 1 in it. If n != 1 then the previous answer holds.

Answer 2

Let:

G(m) Be the number of values i , where 0<=i<=10^m that contain the digit n!=0

As the previous answer says, when n!=0 , the answer is:

G(m) = 10^m - 9^m

So, G(16) = 10,000,000,000,000,000 - 1,853,020,188,851,841 = 8,126,979,811,148,159

When n=0 , there are two possible ways to look at it.

1) Only digits after the first non-zero digit are counted 2) All m digits are counted, even if the number starts with many 0 digits

For the second case, then the answer when n=0 is the same as when n!=0 .

F(m) Be the number of values i , where 0<=i<=10^m that contain the digit n=0

For the first case, then the answer is a little more complicated. In an m digit number, the first digit is not 0, but the remaining m-1 digits can be anything. So, G(m-1) of the numbers defined by those m-1' digits can contain a zero. Since there are 9 (non zero) possibilities for the leading digit, m-1' digits can contain a zero. Since there are 9 (non zero) possibilities for the leading digit, F(m) = 9*G(m-1)`.

Let H(m,n) Be the number of values i , where 0<=i<=10^m that contain the digit n

Then:

H(m,n) = G(m) if n!=0 , and 9*G(m-1) if n=0

where

G(m) = 10^m - 9^m

Thus, H(16,n) = 8,126,979,811,148,159 if n!=0 and H(16,0) = 7,146,979,811,148,159