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Dream

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

p是一個質數，由費馬小定理，a∈Z，所以a^p≡ a mod p

費馬小定理：假如p是質數，且gcd(a,p)=1，那麽 a(p-1)≡1（mod p），例如：假如a是整數，p是質數，則a,p顯然互質(即兩者只有一個公約數1)，那麽我們可以得到費馬小定理的一個特例，即當p為質數時候， a^(p-1)≡1(mod p)。

所以對於0≤x，y≤p，

所以將m+n定義為(m+n)%p

將m*n定義為m*n%p

易證滿足集合相等的約束

#include "bits/stdc++.h"

using namespace std;

const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
#define ull unsigned long long
#define ll long long

int main() {
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    int t, p;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &p);
        int flag;
        for (int i = 0; i < p; i++) {
            flag = 0;
            for (int j = 0; j < p; j++) {
                if (flag) cout << " ";
                cout << (i + j) % p ;
                flag = 1;

            }
            cout<<endl;
        }
        for (int i = 0; i < p; i++) {
            flag = 0;
            for (int j = 0; j < p; j++) {
                if (flag) cout << " ";
                cout << (i * j) % p ;
                flag = 1;

            }
            cout<<endl;
        }
    }
    return 0;
}

View Code

2018 CCPC網絡賽 Dream （費馬小定理）