编程实现显示如下形式的数字矩阵：

1 0 0 0 0 0
2 1 0 0 0 0
3 2 1 0 0 0
4 3 2 1 0 0
5 4 3 2 1 0
6 5 4 3 2 1

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
using namespace std;

int main(){
	int i,j,t,a[6][6]={{0,0,0,0,0,0},{0,0,0,0,0,0},{0,0,0,0,0,0},{0,0,0,0,0,0},{0,0,0,0,0,0},{0,0,0,0,0,0}};
	for(i=0;i<6;i++){
		t=i+1;
		for(j=0;j<=i;j++)
			a[i][j]=t--;
	}
	for(i=0;i<6;i++){
		for(j=0;j<6;j++)
			printf("%d ",a[i][j]);
		printf("\n");
	}
	system("pause");
    return 0;
}

• 将a[3][3]按顺时针方向旋转90度。

0	  2	    9
5	  13    6
27    11	1

转变为

27    5	  0
11   13	  2
1	  6	  9

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
using namespace std;

int main(){
	int i,j;
	int a[3][3]={{0,2,9},{5,13,6},{27,11,1}};
    int b[3][3]={0,0,0,0,0,0,0,0,0};
	for(i=0;i<3;i++)
		for(j=0;j<3;j++)
		    b[i][2-j]=a[j][i];
	for(i=0;i<3;i++){
		for(j=0;j<3;j++)
			printf("%d ",b[i][j]);
		printf("\n");
	}
	system("pause");
	return 0;
}