zoj 3212 K-Nice（数学题）

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3212

转载请注明出处：http://blog.csdn.net/u012860063

K-Nice Time Limit: 1 Second       Memory Limit: 32768 KB       Special Judge

This is a super simple problem. The description is simple, the solution is simple. If you believe so, just read it on. Or if you don't, just pretend that you can't see this one.

We say an element is inside a matrix if it has four neighboring elements in the matrix (Those at the corner have two and on the edge have three). An element inside a matrix is called "nice" when its value equals the sum of its four neighbors. A matrix is called "k-nice" if and only if k of the elements inside the matrix are "nice".

Now given the size of the matrix and the value of k, you are to output any one of the "k-nice" matrix of the given size. It is guaranteed that there is always a solution to every test case.

Input

The first line of the input contains an integer T (1 <= T <= 8500) followed by T test cases. Each case contains three integers n, m, k (2 <= n, m <= 15, 0 <= k <= (n - 2) * (m - 2)) indicating the matrix size n * m and it the "nice"-degree k.

Output

For each test case, output a matrix with n lines each containing m elements separated by a space (no extra space at the end of the line). The absolute value of the elements in the matrix should not be greater than 10000.

Sample Input

2
4 5 3
5 5 3

Sample Output

2 1 3 1 1
4 8 2 6 1
1 1 9 2 9
2 2 4 4 3
0 1 2 3 0
0 4 5 6 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0

Author:  ZHUANG, Junyuan

Source: The 6th Zhejiang Provincial Collegiate Programming Contest

题意：

任意输出一个矩阵要求其中有且仅有K个满足题意的点！

满足要求的点即要此点的上下左右的和等于此点！

代码如下：

#include<cstdio>
int map[22][22];
void bai(int x,int y)
{
	map[x][y+1]=0;
	map[x][y-1]=0;
	map[x+1][y]=0;
	map[x-1][y]=0;
}
int main()
{
	int t,x,y,k,i,j;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d%d",&x,&y,&k);
		for(i=0;i<=x+1;i++)
		{
			for(j=0;j<=y+1;j++)
				map[i][j]=1;
		}
		int count=0,flag=0;
		if(k!=0)
		{
		for(i=2;i<=x-1;i++)
		{
			for(j=2;j<=y-1;j++)
			{
				map[i][j]=0;
				bai(i,j);
				count++;
				if(count==k)
				{
					flag=1;
					break;
				}
			}
			if(flag==1)
				break;
		}
		}
		for(i=1;i<=x;i++)
		{
			
			printf("%d",map[i][1]);
			for(j=2;j<=y;j++)
			{
				
				printf(" %d",map[i][j]);
			}
			printf("\n");
		}
	}
	return 0;
}