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<>矩阵的舞蹈
Problem Description
矩阵是非常美妙的东西，可以用来解方程，以及解决一些图论的问题等，应用很广泛。即使没有学过线性代数，大家也一定接触过矩阵，在编程中可以理解为二维的表。
矩阵有很多操作就像舞蹈一样，如行列的置换，矩阵的转置等。今天我们只看矩阵的旋转，希望得到当前矩阵顺时针旋转90度以后得到的矩阵。
Input
输入数据的第一行是一个正整数T，代表有T组测试样例。接下来T组数据，每组数据第一行是两个整数M,N (0 < M , N <
100),分别代表矩阵的行数和列数。然后是矩阵本身，共M行，每行N个数据用空格隔开。
Output
对于每组输入的矩阵，第一行输出Case #k:（k为该组数据的序号,具体格式见样例），然后输出其旋转后的矩阵。
Sample Input
2
4 4
1 2 3 4
5 6 7 8
6 6 6 6
8 8 8 8
2 3
1 2 3
4 5 6
Sample Output
Case #1:
8 6 5 1
8 6 6 2
8 6 7 3
8 6 8 4
Case #2:
4 1
5 2
6 3
示例代码：
#include<stdio.h> int main(void) { int N, i, j; int m, n, a[100][100];
scanf("%d", &N); for (int I = 1; I <= N; I++) { scanf("%d %d", &m, &n); for (i
= 0; i < m; i++) for (j = 0; j < n; j++) scanf("%d", &a[i][j]); printf("Case
#%d:\n", I); for(i=0;i<n;i++) for (j = 0; j < m; j++) { if(j==m-1)
printf("%d\n", a[m - 1 - j][i]); else printf("%d ", a[m - 1 - j][i]); } }
return 0; }
