原题：
Description
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 byT test cases. Each case contains three integers n, m, k (2 <= n, m <= 15, 0 <= k <= (n - 2) * (m - 2)) indicating the matrix sizen * 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
分析：
先赋值，全部赋值为1，然后上下左右判断————按照要求加0.
原码：
#include<stdio.h> int main() { int t,n,m,k,a[15][15]; scanf("%d",&t); while(t--) { int sum=0; scanf("%d%d%d",&n,&m,&k); for(int i=0; i<n; i++) { for(int j=0; j<m; j++) { a[i][j]=1; } } for(int i=1; i<n-1; i++) { for(int j=1; j<m-1; j++) { if(sum==k) break; else { a[i][j]=0; a[i-1][j]=0; a[i+1][j]=0; a[i][j-1]=0; a[i][j+1]=0; sum++; } } } for(int i=0; i<n; i++) { for(int j=0; j<m; j++) { if(j==0) printf("%d",a[i][j]); else printf(" %d",a[i][j]); } printf("\n"); } } return 0; }
