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

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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

构造一个n*n的矩阵，保证每行每列的值都不相同。

通过n == 2，可以推出n == 3的时候，需要新加入两个数，但是添加1构成3就没法构成-2，添加-1构成-2就得不到3，以此类推n为奇数的情况都是不可以的，n为偶数时，只需要构造每行每列的1跟-1的值不一样就OK了，可以先空出左下对角线，上三角填充1，下三角填充0，然后对角线0，1平分即可。

代码实现：

/*
Look at the star
Look at the shine for U
*/ 

#include<bits/stdc++.h>
#define sl(x) scanf("%lld",&x)
using namespace std;
typedef long long ll;
const int N = 1e6+5;
const ll mod = 1e9+7;
const ll INF = 1e18;
int s[205][205];
int main()
{
	int n,i,j,k,t;
	scanf("%d",&t);
	while(t--)
	{
		memset(s,0,sizeof(s));
		scanf("%d",&n);
		if(n&1) puts("impossible");
		else
		{
			puts("possible");
			for(i = 0;i < n-1;i++)
				for(j = 0;j < n-i-1;j++)
					s[i][j] = 1;
					
			for(i = 1;i < n;i++)
				for(j = n-1;j >= n-i;j--)
					s[i][j] = -1;
					
			for(i = n-1;i >= n/2;i--) s[i][n-i-1] = 1;
			for(i = 0;i < n;i++)
			{
				for(j = 0;j < n;j++)
				{
					if(j) printf(" "); printf("%d",s[i][j]);
				}
				puts("");
			}
		}
	}
	
}