Dream
Special Judge
 

Problem Description

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=mp+np, where m,n,p are real numbers. Let's call it ``Beginner's Dream''.

For instance, (1+4)2=52=25, but 12+42=17≠25. Moreover, 9+16−−−−−√=25−−√=5, which does not equal 3+4=7. 

Fortunately, in some cases when p is a prime, the identity

(m+n)p=mp+np

holds true for every pair of non-negative integers m,n which are less than p, 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=mp+np is a valid identity for all non-negative integers m,n less than p. Power is defined as

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) to make the set {qk|0<k<p,k∈Z} equal to {k|0<k<p,k∈Z}. What's more, the set of non-negative integers less than p ought to be closed under the operation of your definitions.

Hint

Hint for sample input and output:
From the table we get 0+1=1, and thus (0+1)2=12=1⋅1=1. On the other hand, 02=0⋅0=0, 12=1⋅1=1, 02+12=0+1=1.
They are the same.

Input

The first line of the input contains an positive integer T(T≤30) indicating the number of test cases.

For every case, there is only one line contains an integer p(p<210), described in the problem description above. p is guranteed to be a prime.

Output

For each test case, you should print 2p lines of p integers.

The j-th(1≤j≤p) integer of i-th(1≤i≤p) line denotes the value of (i−1)+(j−1). The j-th(1≤j≤p) integer of (p+i)-th(1≤i≤p) line denotes the value of (i−1)⋅(j−1).

Sample Input

1 2

Sample Output

0 1 1 0 0 0 0 1

Source

2018中国大学生程序设计竞赛 - 网络选拔赛

 题意：

题目很难读懂 ， 给你一个质数p ， 得出的答案是一个p * p 的矩阵

让你重新定义  ‘+’   和   ‘*’   使得（m+n)^p = m^p + n^p;(其中 m , n 指的是小于p的非负整数 ）

思路：

首先要知道费马小定理的内容：

当p为质数时候， a^(p-1)≡1(mod p)。

（如果不是很理解 ， 可以看下这个费马小定理的应用：

例如计算 除以13的余数

故余数为3。）

下面代码：

#include <bits/stdc++.h>
using namespace std;
int main()
{
	int t , p;
	scanf("%d" , &t);
	while(t--)
	{
		scanf("%d" , &p);
		for(int i = 0 ; i < p ; i++)
		{
			for(int j = 0 ; j < p ; j++)
			{
				printf("%d " , (i+j)%p);
			}
			printf("\n");
		}
	
		for(int i = 0 ; i < p ; i++)
		{
			for(int j = 0 ; j < p ; j++)
			{
				printf("%d " , (i*j)%p);
			}
				printf("\n");
		}
	}
	return 0;
 }