10748: ABC137 - F - Polynomial Construction
[Creator : ]
Description
Given are a prime number $p$ and a sequence of $p$ integers $a_0, \ldots, a_{p-1}$ consisting of zeros and ones.
Find a polynomial of degree at most $p-1$, $f(x) = b_{p-1} x^{p-1} + b_{p-2} x^{p-2} + \ldots + b_0$, satisfying the following conditions:
- For each $i$ $(0 \leq i \leq p-1)$, $b_i$ is an integer such that $0 \leq b_i \leq p-1$.
- For each $i$ $(0 \leq i \leq p-1)$, $f(i) \equiv a_i \pmod p$.
Find a polynomial of degree at most $p-1$, $f(x) = b_{p-1} x^{p-1} + b_{p-2} x^{p-2} + \ldots + b_0$, satisfying the following conditions:
- For each $i$ $(0 \leq i \leq p-1)$, $b_i$ is an integer such that $0 \leq b_i \leq p-1$.
- For each $i$ $(0 \leq i \leq p-1)$, $f(i) \equiv a_i \pmod p$.
Input
Input is given from Standard Input in the following format:
```
$p$
$a_0$ $a_1$ $\ldots$ $a_{p-1}$
```
```
$p$
$a_0$ $a_1$ $\ldots$ $a_{p-1}$
```
Output
Print $b_0, b_1, \ldots, b_{p-1}$ of a polynomial $f(x)$ satisfying the conditions, in this order, with spaces in between.
It can be proved that a solution always exists. If multiple solutions exist, any of them will be accepted.
It can be proved that a solution always exists. If multiple solutions exist, any of them will be accepted.
Constraints
- $2 \leq p \leq 2999$
- $p$ is a prime number.
- $0 \leq a_i \leq 1$
- $p$ is a prime number.
- $0 \leq a_i \leq 1$
Sample 1 Input
2
1 0
Sample 1 Output
1 1
$f(x) = x + 1$ satisfies the conditions, as follows:
- $f(0) = 0 + 1 = 1 \equiv 1 \pmod 2$
- $f(1) = 1 + 1 = 2 \equiv 0 \pmod 2$
Sample 2 Input
3
0 0 0
Sample 2 Output
0 0 0
$f(x)=0$ is also valid.
Sample 3 Input
5
0 1 0 1 0
Sample 3 Output
0 2 0 1 3