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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
2 1 2
输出
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("");
}
}
}
}