牛客网暑期ACM多校训练营（第四场）Another Distinct Values

链接：https://www.nowcoder.com/acm/contest/142/D
来源：牛客网
 

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 131072K，其他语言262144K
Special Judge, 64bit IO Format: %lld

题目描述

Chiaki has an n x n matrix. She would like to fill each entry by -1, 0 or 1 such that r1,r2,...,rn,c1,c2, ..., cn are distinct values, where ri be the sum of the i-th row and ci be the sum of the i-th column.

输入描述:

There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 200), indicating the number of test cases. For each test case:
The first line contains an integer n (1 ≤ n ≤ 200) -- the dimension of the matrix.

输出描述:

For each test case, if no such matrix exists, output ``impossible'' in a single line. Otherwise, output ``possible'' in the first line. And each of the next n lines contains n integers, denoting the solution matrix. If there are multiple solutions, output any of them.

示例1

输入

复制

2
1
2

输出

复制

impossible
possible
1 0
1 -1

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<string>
#define long long LL
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define lson o<<1
#define rson o<<1|1
using namespace std;
int main()
{
	int n;
	int t;
	cin >> t;
	while(t--)
	{
		cin >> n;
		if(n & 1)
			puts("impossible");
		else{
			puts("possible");
			for(int i = 1;i <= n; i++)
			{
				for(int j = 1;j <= n; j++)
				{
					if(i == j)
						cout << i%2 << " ";
					else if(i < j)
						cout << -1 << " ";
					else
						cout << 1 << " ";
				}
				cout << endl; 
			}
		}
	}
	
	
    return 0;
}