English · Original
Chinese · Translation
Formal · Original
A progress bar is an element of graphical interface that displays the progress of a process for this very moment before it is completed. Let's take a look at the following form of such a bar.
A bar is represented as _n_ squares, located in line. To add clarity, let's number them with positive integers from 1 to _n_ from the left to the right. Each square has saturation (_a__i_ for the _i_\-th square), which is measured by an integer from 0 to _k_. When the bar for some _i_ (1 ≤ _i_ ≤ _n_) is displayed, squares 1, 2, ... , _i_ - 1 has the saturation _k_, squares _i_ + 1, _i_ + 2, ... , _n_ has the saturation 0, and the saturation of the square _i_ can have any value from 0 to _k_.
So some first squares of the progress bar always have the saturation _k_. Some last squares always have the saturation 0. And there is no more than one square that has the saturation different from 0 and _k_.
The degree of the process's completion is measured in percents. Let the process be _t_% completed. Then the following inequation is fulfilled:
An example of such a bar can be seen on the picture.
<center></center>For the given _n_, _k_, _t_ determine the measures of saturation for all the squares _a__i_ of the progress bar.
## Input
We are given 3 space-separated integers _n_, _k_, _t_ (1 ≤ _n_, _k_ ≤ 100, 0 ≤ _t_ ≤ 100).
## Output
Print _n_ numbers. The _i_\-th of them should be equal to _a__i_.
[samples]
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**Definitions**
Let $ n, k, t \in \mathbb{Z} $ be given integers, where:
- $ n $: number of squares in the progress bar,
- $ k $: maximum saturation value,
- $ t $: completion percentage ($ 0 \leq t \leq 100 $).
Let $ A = (a_1, a_2, \dots, a_n) $ be the sequence of saturation values for the $ n $ squares, where each $ a_i \in \{0, 1, \dots, k\} $.
**Constraints**
1. $ 1 \leq n \leq 100 $
2. $ 1 \leq k \leq 100 $
3. $ 0 \leq t \leq 100 $
**Objective**
Compute $ A = (a_1, \dots, a_n) $ such that:
- There exists an index $ i \in \{1, \dots, n\} $ where:
- $ a_j = k $ for all $ j < i $,
- $ a_j = 0 $ for all $ j > i $,
- $ a_i \in \{0, 1, \dots, k\} $ (at most one such $ i $ with intermediate value).
The index $ i $ and value $ a_i $ are determined by the percentage $ t $:
Let $ p = \frac{t}{100} \cdot n $ be the fractional progress position.
Define:
- $ i = \lfloor p \rfloor + 1 $,
- $ \text{frac} = p - \lfloor p \rfloor $,
- $ a_i = \lfloor k \cdot \text{frac} + 0.5 \rfloor $ (rounded to nearest integer).
Then:
$$
a_j =
\begin{cases}
k & \text{if } j < i, \\
\lfloor k \cdot (p - \lfloor p \rfloor) + 0.5 \rfloor & \text{if } j = i, \\
0 & \text{if } j > i.
\end{cases}
$$
If $ p \geq n $, then $ i = n $ and $ a_n = k $, with all $ a_j = k $.
If $ p < 1 $, then $ i = 1 $ and $ a_1 = \lfloor k \cdot p + 0.5 \rfloor $, with $ a_j = 0 $ for $ j > 1 $.
API Response (JSON)
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