Interesting Function (Codeforces 1538F)
Tags: Math, binary search

You are given two integers $$$l$$$ and $$$r$$$, where $$$l < r$$$. We will add $$$1$$$ to $$$l$$$ until the result is equal to $$$r$$$. Thus, there will be exactly $$$r-l$$$ additions performed. For each such addition, let's look at the number of digits that will be changed after it.For example:   if $$$l=909$$$, then adding one will result in $$$910$$$ and $$$2$$$ digits will be changed;  if you add one to $$$l=9$$$, the result will be $$$10$$$ and $$$2$$$ digits will also be changed;  if you add one to $$$l=489999$$$, the result will be $$$490000$$$ and $$$5$$$ digits will be changed. Changed digits always form a suffix of the result written in the decimal system.Output the total number of changed digits, if you want to get $$$r$$$ from $$$l$$$, adding $$$1$$$ each time.

Input: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$). Then $$$t$$$ test cases follow.Each test case is characterized by two integers $$$l$$$ and $$$r$$$ ($$$1 \le l < r \le 10^9$$$).

Output: For each test case, calculate the total number of changed digits if you want to get $$$r$$$ from $$$l$$$, adding one each time.


Sample Inputs:
4
1 9
9 10
10 20
1 1000000000

Sample Outputs:
8
2
11
1111111110

Solution (Python):
for s in [*open(0)][1:]:
    (x, y) = map(int, s.split())
    r = 0
    while y:
        r += y - x
        x //= 10
        y //= 10
    print(r)

Solution (Python):
for s in [*open(0)][1:]:
    (x, y) = map(int, s.split())
    r = 0
    while y:
        r += y - x
        x //= 10
        y //= 10
    print(r)

Solution (Python):
t = int(input())
result = []
for _ in range(t):
    (l, r) = map(int, input().split())
    i = 1000000000
    r_count = 0
    l_count = 0
    while i > 0:
        r_count += r // i
        l_count += l // i
        i //= 10
    result.append(r_count - l_count)
print(*result, sep='\n')