Yet Another Two Integers Problem (Codeforces 1409A)
Tags: Greedy, math

You are given two integers $$$a$$$ and $$$b$$$.In one move, you can choose some integer $$$k$$$ from $$$1$$$ to $$$10$$$ and add it to $$$a$$$ or subtract it from $$$a$$$. In other words, you choose an integer $$$k \in [1; 10]$$$ and perform $$$a := a + k$$$ or $$$a := a - k$$$. You may use different values of $$$k$$$ in different moves.Your task is to find the minimum number of moves required to obtain $$$b$$$ from $$$a$$$.You have to answer $$$t$$$ independent test cases.

Input: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. Then $$$t$$$ test cases follow.The only line of the test case contains two integers $$$a$$$ and $$$b$$$ ($$$1 \le a, b \le 10^9$$$).

Output: For each test case, print the answer: the minimum number of moves required to obtain $$$b$$$ from $$$a$$$.

Note: In the first test case of the example, you don't need to do anything.In the second test case of the example, the following sequence of moves can be applied: $$$13 \rightarrow 23 \rightarrow 32 \rightarrow 42$$$ (add $$$10$$$, add $$$9$$$, add $$$10$$$).In the third test case of the example, the following sequence of moves can be applied: $$$18 \rightarrow 10 \rightarrow 4$$$ (subtract $$$8$$$, subtract $$$6$$$).

Sample Inputs:
6
5 5
13 42
18 4
1337 420
123456789 1000000000
100500 9000

Sample Outputs:
0
3
2
92
87654322
9150

Solution (Python):
def main():
    for s in [*open(0)][1:]:
        a = [*map(int, s.split())]
        b = abs(a[0] - a[1])
        if b % 10:
            print(b // 10 + 1)
        else:
            print(b // 10)
main()

Solution (Python):
t = int(input())
min_ls = []
for i in range(t):
    (a, b) = (int(j) for j in input().split())
    min_ls.append(str(abs(a - b) // 10 + (1 if abs(a - b) % 10 != 0 else 0)))
print('\n'.join(min_ls))

Solution (Python):
import math
t = int(input())
res = []
while t:
    t -= 1
    inp = str(input()).split(' ')
    (a, b) = (int(inp[0]), int(inp[1]))
    res.append(math.ceil(abs(a - b) / 10))
for r in res:
    print(r)