Rectangular Congruence

题意翻译

给定一个质数 $n$（$n \leq 350$） 和一个序列 $b_1, b_2, ..., b_n$（对于 $\forall i$ 有 $0 \leq b_i < n$），你需要构造一个 $n \times n$ 的矩阵 $a$，满足： - 对于 $\forall i, j \leq n$ 有 $0 \leq a_{i, j} < n $。 - 对于 $\forall 1 \leq r_1 \leq r_2 \leq n, 1 \leq c_1 \leq c_2 \leq n$ 有 $a_{r1, c1} + a_{r2, c2} \not\equiv a_{r1, c2} + a_{r2, c1} \pmod n $。 - 对于 $\forall 1 \leq i \leq n, 1 \leq c_1 \leq c_2 \leq n$ 有 $ a_{i, i} = b_i $。

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.

输入输出格式

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

输入样例 #1

2
0 0

输出样例 #1

0 1 
0 0

输入样例 #2

3
1 1 1

输出样例 #2

1 2 2
1 1 0
1 0 1

输入样例 #3

5
1 4 1 2 4

输出样例 #3

1 0 1 3 4
1 4 3 1 0
2 4 1 0 2
1 2 2 2 2
2 2 0 1 4

说明

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.