梦开始的地方(Dream)

英文题面：

题目出处:Vjudge
Freshmen frequently make an error in computing the power of a sum of real numbers, which usually origins from an incorrect equation ( m + n ) p = m p + n p (m+n)^p = m ^p + n ^ p (m+n)p=mp+np
, where m, n, pm,n,p are real numbers. Let’s call it ``Beginner’s Dream’'.

For instance, ( 1 + 4 ) 2 = 5 2 = 25 (1+4)^2 = 5^2 = 25 (1+4)2=52=25 but 1 2 + 4 2 = 17 ≠ 25 1^2 + 4 ^2 = 17 \ne 25 12+42=17=25
Moreover, 9 + 16 = 25 = 5 \sqrt{9+16}=\sqrt{25}=5 9+16 =25 =5 which does not equal 3 + 4 = 7 3+4=7 3+4=7.
Fortunately, in some cases when pp is a prime, the identity
( m + n ) p = m p + n p (m+n)^p = m^p + n^p (m+n)p=mp+np
holds true for every pair of non-negative integers m, nm,n which are less than pp, with appropriate definitions of addition and multiplication.

You are required to redefine the rules of addition and multiplication so as to make the beginner’s dream realized.

Specifically, you need to create your custom addition and multiplication, so that when making calculation with your rules the equation ( m + n ) p = m p + n p (m+n)^p=m^p+n^p (m+n)p=mp+np is a valid identity for all non-negative integers m,nm,n less than pp. Power is defined as
a p = { 1 , p = 0 a p − 1 ⋅ a , p > 0 a^p = \left\{ \begin{array}{ll} 1, & p = 0 \\ a^{p-1} \cdot a, & p > 0 \end{array} \right. ap={1,ap−1⋅a,p=0p>0

Obviously there exists an extremely simple solution that makes all operation just produce zero. So an extra constraint should be satisfied that there exists an integer q:(0 < q < p)q(0<q<p) to make the set { q k ∣ 0 < k < p , k ∈ Z } q k ∣ 0 < k < p , k ∈ Z e q u a l t o { k ∣ 0 < k < p , k ∈ Z } k ∣ 0 < k < p , k ∈ Z . \{ q^k \,|\, 0 < k < p, k \in \mathbb{Z} \}{q^k ∣0<k<p,k∈Z} equal \ to \{k\,|\,0<k<p,k \in \mathbb{Z}\}{k∣0<k<p,k∈Z}. {qk∣0<k<p,k∈Z}qk∣0<k<p,k∈Zequal to{k∣0<k<p,k∈Z}k∣0<k<p,k∈Z. What’s more, the set of non-negative integers less than pp ought to be closed under the operation of your definitions.

描述

祥祥经常在计算实数和的幂时犯错误，这通常源于一个错误的等式 ( m + n ) p = m p + n p (m+n)^p=m^p+n^p (m+n)p=mp+np，其中 m , n , p m,n,p m,n,p 是实数。我们把它称为“祥祥的梦想”。

例如 ( 1 + 4 ) 2 = 5 2 = 25 ≠ 1 2 + 4 2 (1+4)^2 = 5^2=25\neq 1^2 + 4^2 (1+4)2=52=25=12+42 ，又如 9 + 16 = 25 = 5 ≠ 3 + 4 \sqrt{9 + 16} = \sqrt{25} = 5 \neq 3 + 4 9+16 =25 =5=3+4 .

幸运的是，在一些情况下，当 p p p 是质数的时候，如果重新定义加法、乘法，等式 ( m + n ) p = m p + n p (m+n)^p=m^p+n^p (m+n)p=mp+np 对于任何小于 p p p 的非负整数 m , n m,n m,n 都成立。

你需要重新定义加法和乘法，来实现祥祥的梦想。

更具体的说，你需要重新定义加法和乘法来使得等式 ( m + n ) p = m p + n p (m+n)^p=m^p+n^p (m+n)p=mp+np 对于任何小于 p p p 的非负整数 m , n m,n m,n 都成立。下面是幂的定义：
a p = { 1 , p = 0 a p − 1 ⋅ a , p > 0 a^p = \left\{ \begin{array}{ll} 1, & p = 0 \\ a^{p-1} \cdot a, & p > 0 \end{array} \right. ap={1,ap−1⋅a,p=0p>0

显然，这里存在一个非常极端的解，就是让所有的运算结果都是 0 0 0 ，为了避免这种情况，你需要让你的解满足：存在一个正整数$ q (0 < q < p)$ 使得集合 { q k ∣ 0 < k < p , k ∈ Z } \{q^k|0<k<p, k\in \mathbb{Z}\} {qk∣0<k<p,k∈Z} 和集合 { k ∣ 0 < k < p , k ∈ Z } \{k|0<k<p, k\in \mathbb{Z}\} {k∣0<k<p,k∈Z} 相等。

注意，你定义的运算需要让小于 p p p 的非负整数集对其封闭。

输入描述

输入占一行，是一个正质数 p p p ，意义见上述内容。

输出描述

输出占 2 p 2p 2p 行，每行是 p p p 个整数。

第 i i i 行（ 1 ≤ i ≤ p 1\leq i\leq p 1≤i≤p）的第 j j j 个（ 1 ≤ j ≤ p 1\leq j\leq p 1≤j≤p）整数表示 ( i − 1 ) + ( j − 1 ) (i-1)+(j-1) (i−1)+(j−1) 的值。第 ( p + i ) (p+i) (p+i) 行（ 1 ≤ i ≤ p 1\leq i\leq p 1≤i≤p）的第 j j j 个（ 1 ≤ j ≤ p 1\leq j\leq p 1≤j≤p）整数表示 ( i − 1 ) ⋅ ( j − 1 ) (i-1)\cdot(j-1) (i−1)⋅(j−1) 的值。

提示

0 < p < 2 10 0 < p < 2^{10} 0<p<210

样例

样例输入

样例输出

题解代码

题目的所有条件都满足模p剩余域(即模P剩余系)的条件
证明：
首先，题目中的 ( m + n ) p = m p + n p ( p ∈ 素数 ) (m + n)^p = m^p + n^p(p \in 素数) (m+n)p=mp+np(p∈素数)，要满足一个条件，即：
( m + n ) p ≡ m p + n p ( m o d p ) (m + n) ^p \equiv m^p + n^p \pmod{p} (m+n)p≡mp+np(modp)
将二项式展开，得到：
( m + n ) p = ∑ i = 0 p ( p i ) m n − i n i ≡ m p + n p ( m o d p ) (m + n) ^p = \sum_{i = 0}^{p} \dbinom{p}{i} m^{n - i}n^i \equiv m^p + n^p \pmod{p} (m+n)p=i=0∑p(ip)mn−ini≡mp+np(modp)
同时，在模P剩余系 ( p ∈ 素数 ) (p \in 素数) (p∈素数)中， ( p k ) ≡ 0 ( m o d p ) \dbinom{p}{k} \equiv 0 \pmod{p} (kp)≡0(modp)
简单的证明一下：
首先： ( p k ) = p ! k ! ( p − k ) ! \dbinom{p}{k} = \frac{p!}{k!(p-k)!} (kp)=k!(p−k)!p!
∴ 原式 = ∏ i = p p − k + 1 i k ! \therefore 原式 = \frac{\prod_{i = p}^{p - k + 1} i}{k!} ∴原式=k!∏i=pp−k+1i
∵ k ∈ 模 p 剩余系 \because k \in 模p剩余系 ∵k∈模p剩余系
k 模 p 的逆元也 ∈ 模 p 剩余系 k模p 的逆元也 \in 模p剩余系 k模p的逆元也∈模p剩余系
注：逆元
∴ ∏ i = p p − k + 1 i k ! = ∏ i = p p − k + 1 i × ∏ i = k 1 k 模 p 的逆元 ≡ 0 ( m o d p ) \therefore \frac{\prod_{i = p}^{p - k + 1} i}{k!} = \prod_{i = p}^{p - k + 1} i \times \prod_{i = k}^{1}k模p的逆元 \equiv 0 \pmod{p} ∴k!∏i=pp−k+1i=i=p∏p−k+1i×i=k∏1k模p的逆元≡0(modp)
证毕。
所以，一切只要按模P剩余系的加乘来就好了(即正常 R R R内的加乘)。
代码：

#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 5;
const int M = 1005;
typedef long long ll;
signed main(){int p;scanf("%d",&p);for(int i = 1 ; i <= p ; i ++) {for (int j = 1 ; j <= p ; j ++) {printf("%d " , ((i - 1) + (j - 1)) % p) % p;}printf("\n");}for (int i = 1 ; i <= p ; i ++) {for (int j = 1 ; j <= p ; j ++) {printf("%d " , ((i - 1) % p * (j - 1) % p) %p);}printf("\n");}return 0;
}