Round Down the Price (Codeforces 1702A)
Tags: Others, constructive algorithms

At the store, the salespeople want to make all prices round. In this problem, a number that is a power of $$$10$$$ is called a round number. For example, the numbers $$$10^0 = 1$$$, $$$10^1 = 10$$$, $$$10^2 = 100$$$ are round numbers, but $$$20$$$, $$$110$$$ and $$$256$$$ are not round numbers. So, if an item is worth $$$m$$$ bourles (the value of the item is not greater than $$$10^9$$$), the sellers want to change its value to the nearest round number that is not greater than $$$m$$$. They ask you: by how many bourles should you decrease the value of the item to make it worth exactly $$$10^k$$$ bourles, where the value of $$$k$$$ — is the maximum possible ($$$k$$$ — any non-negative integer).For example, let the item have a value of $$$178$$$-bourles. Then the new price of the item will be $$$100$$$, and the answer will be $$$178-100=78$$$.

Input: The first line of input data contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases .Each test case is a string containing a single integer $$$m$$$ ($$$1 \le m \le 10^9$$$) — the price of the item.

Output: For each test case, output on a separate line a single integer $$$d$$$ ($$$0 \le d < m$$$) such that if you reduce the cost of the item by $$$d$$$ bourles, the cost of the item will be the maximal possible round number. More formally: $$$m - d = 10^k$$$, where $$$k$$$ — the maximum possible non-negative integer.

Note: In the example:  $$$1 - 0 = 10^0$$$,  $$$2 - 1 = 10^0$$$,  $$$178 - 78 = 10^2$$$,  $$$20 - 10 = 10^1$$$,  $$$999999999 - 899999999 = 10^8$$$,  $$$9000 - 8000 = 10^3$$$,  $$$987654321 - 887654321 = 10^8$$$. Note that in each test case, we get the maximum possible round number.

Sample Inputs:
7
1
2
178
20
999999999
9000
987654321

Sample Outputs:
0
1
78
10
899999999
8000
887654321

Solution (Python):
out = []
for i in range(int(input())):
    a = int(input())
    b = len(str(a))
    out.append(a - 10 ** (b - 1))
print(*out, sep='\n')

Solution (Python):
for m in [*open(0)][1:]:
    print(int(m) - 10 ** (len(m) - 2))

Solution (Python):
for m in [*open(0)][1:]:
    print(int(m) - 10 ** (len(m) - 2))