codeforces#P1982B. Collatz Conjecture
Collatz Conjecture
Description
Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:
There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:
- increase $x$ by $1$, then
- while the number $x$ is divisible by $y$, divide it by $y$.
For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.
Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.
Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.
The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.
For each test case, output a single integer — the number obtained after applying $k$ operations.
Input
Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.
The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.
Output
For each test case, output a single integer — the number obtained after applying $k$ operations.
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2
Note
In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.
In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.
In the third test case, $x$ changes as follows:
- After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.
- After the second operation, $x = 2$.
- After the third operation, $x = 3$.
- After the fourth operation, $x = 4$.
- After the fifth operation, $x = 1$.