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最新推荐文章于 2019-08-25 00:44:05 发布

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最新推荐文章于 2019-08-25 00:44:05 发布

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分类专栏：杭电oj练习文章标签： CCPC

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博客围绕解决新手常犯的实数幂运算错误展开，要求在给定素数p的情况下，重新定义加法和乘法规则，使(m+n)^p=m^p+n^p对所有小于p的非负整数m、n成立。依据费马小定理，在模p意义下该等式恒成立，且运算封闭。

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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

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chendu

给定的p是素数，要求给定一个加法运算表和乘法运算表，

使(m+n)p=mp+np(0≤m,n<p)
因为给定的p是素数，根据费马小定理得 (m+n)p−1≡1(mod p)
因此，(m+n)p≡m+n (mod p)
同时，mp+np≡m+n (mod p)。
所以在模p意义下，(m+n)p=mp+np(0≤m,n<p)恒成立，且加法运算与乘法运算封闭。

#include<bits/stdc++.h>
using namespace std;
const int maxn =1e5+5;
typedef long long LL;

int main()
{

    int p;
    int T; scanf("%d",&T);
    while(T--){
        scanf("%d",&p);
        for(int i=0;i<p;++i){
            for(int j=0;j<p;++j){
                printf("%d%c",(i+j)%p,j==p-1?'\n':' ');
            }
        }
        for(int i=0;i<p;++i){
            for(int j=0;j<p;++j){
                printf("%d%c",(i*j%p),j==p-1?'\n':' ');
            }       
        }
    }
    return 0;
}