题目链接:Dream
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1127 Accepted Submission(s): 286
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<2^10), 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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臭长臭长的题目，真的没有看懂题目，后来经大佬点播，好像懂了一点，我好菜。
其实，，只要看懂红色部分，然后在大胆模拟就行了，
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
#include <cstdio>
#include <deque>
#include <vector>
#include <set>
#define pi acos(-1.0)
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
ll p;
scanf("%lld",&p);
for(int i=1;i<=p;i++)
{
for(int j=1;j<=p;j++)
{
printf("%lld ",(i-1+j-1)%p);
}
printf("\n");
}
for(int i=1;i<=p;i++)
{
for(int j=1;j<=p;j++)
{
printf("%lld ",((i-1)*(j-1))%p);
}
printf("\n");
}
}return 0;
}
