D题偶然发现,这个奇数不行,偶数可以,并且都能从上一个可以的情况推出来,可以把答案从n=2开始,然后上面放1下边放-1,然后再调两边,最后看成两个三角形,以对角线为分界线,上面全是1,下面全是-1,对角线上,前n/2行是1,后是0.
#include <iostream>
#include <algorithm>
#include <map>
#include <string>
#include <cstring>
#include <cmath>
#include <stack>
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define ll long long
#define fast ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define sc(a) scanf("%lld",&a)
#define pf(a) printf("%d",a)
#define endl "\n"
#define int long long
#define mem(a,b) memset(a,b,sizeof a)
#define ull unsigned long long
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define lowbit(x) x&(-x)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define x first
#define y second
const double eps = 1e-6;
const int mod = 998244353;
const int MOD = 1e9 + 7;
signed main()
{
int n, t;
sc(t);
while(t--)
{
sc(n);
if(n&1) cout<<"impossible"<<endl;
else
{
cout<<"possible"<<endl;
for(int i=1;i<=n;++i)
{
for(int j=1;j<=n-i;++j)
{
cout<<1<<' ';
}
if(i>n/2) cout<<0<<' ';
else cout<<1<<' ';
for(int j=1;j<i;j++) cout<<-1<<' ';
cout<<endl;
}
}
}
return 0;
}