CF1734E.Rectangular Congruence

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题目描述

You are given a prime number $n$ , and an array of $n$ integers $b_1,b_2,\ldots, b_n$ , where $0 \leq b_i < n$ for each $1 \le i \leq n$ .

You have to find a matrix $a$ of size $n \times n$ such that all of the following requirements hold:

$0 \le a_{i,j} < n$ for all $1 \le i, j \le n$ .
$a_{r_1, c_1} + a_{r_2, c_2} \not\equiv a_{r_1, c_2} + a_{r_2, c_1} \pmod n$ for all positive integers $r_1$ , $r_2$ , $c_1$ , and $c_2$ such that $1 \le r_1 < r_2 \le n$ and $1 \le c_1 < c_2 \le n$ .
$a_{i,i} = b_i$ for all $1 \le i \leq n$ .

Here $x \not \equiv y \pmod m$ denotes that $x$ and $y$ give different remainders when divided by $m$ .

If there are multiple solutions, output any. It can be shown that such a matrix always exists under the given constraints.

输入格式

The first line contains a single positive integer $n$ ( $2 \le n < 350$ ).

The second line contains $n$ integers $b_1, b_2, \ldots, b_n$ ( $0 \le b_i < n$ ) — the required values on the main diagonal of the matrix.

It is guaranteed that $n$ is prime.

输出格式

Print $n$ lines. On the $i$ -th line, print $n$ integers $a_{i, 1}, a_{i, 2}, \ldots, a_{i, n}$ , each separated with a space.

If there are multiple solutions, output any.

输入输出样例

输入#1
```
2
0 0
```
输出#1
```
0 1 
0 0
```
输入#2
```
3
1 1 1
```
输出#2
```
1 2 2
1 1 0
1 0 1
```

输入#3

5
1 4 1 2 4

输出#3

说明/提示

In the first example, the answer is valid because all entries are non-negative integers less than $n = 2$ , and $a_{1,1}+a_{2,2} \not\equiv a_{1,2}+a_{2,1} \pmod 2$ (because $a_{1,1}+a_{2,2} = 0 + 0 \equiv 0 \pmod 2$ and $a_{1,2}+a_{2,1} = 1 + 0 \equiv 1 \pmod 2 $ ). Moreover, the values on the main diagonals are equal to $0,0$ as required.

In the second example, the answer is correct because all entries are non-negative integers less than $n = 3$ , and the second condition is satisfied for all quadruplets $(r_1, r_2, c_1, c_2)$ . For example:

When $r_1=1$ , $r_2=2$ , $c_1=1$ and $c_2=2$ , $a_{1,1}+a_{2,2} \not\equiv a_{1,2}+a_{2,1} \pmod 3$ because $a_{1,1}+a_{2,2} = 1 + 1 \equiv 2 \pmod 3$ and $a_{1,2}+a_{2,1} = 2 + 1 \equiv 0 \pmod 3 $ .
When $r_1=2$ , $r_2=3$ , $c_1=1$ , and $c_2=3$ , $a_{2,1}+a_{3,3} \not\equiv a_{2,3}+a_{3,1} \pmod 3$ because $a_{2,1}+a_{3,3} = 1 + 1 \equiv 2 \pmod 3$ and $a_{2,3}+a_{3,1} = 0 + 1 \equiv 1 \pmod 3 $ .

Moreover, the values on the main diagonal are equal to $1,1,1$ as required.

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