//经典矩阵问题是利用数字生成一个矩阵,而该矩阵刚好是一个正方形,该矩阵内的数字是有
//规律的排序而形成矩阵。比较常见有以下形式
//1.
// 1 2 9 10
// 4 3 8 11
// 5 6 7 12
// 16 15 14 13
//2.
// 1 2 6 7
// 3 5 8 13
// 4 9 12 14
// 10 11 15 16
//3.
// 1 2 3 4
// 12 13 14 5
// 11 16 15 6
// 10 9 8 7
//4.
// 7 6 5 16
// 8 1 4 15
// 9 2 3 14
// 10 11 12 13
#include
using namespace std;
const int N = 5;//可修改
int a[N][N];
void Print(int n)
{
for (int i = 0; i < n; ++i)
{
for (int j = 0; j < n; ++j)
printf("%-2d ",a[i][j]);
cout << endl;
}
cout << endl;
}
void Rectangle1()
{
int i = 0, j = 0;
int lap = 1, m = 1, n;
//cout<< "please input n行数";
//cin >> n;
n = 4;
a[i][j] = m++;
lap++;
while (lap <= n)
{
if (lap % 2 == 0)
{
for (j++; i < lap; i++)
a[i][j] = m++;
i--;
for (j--; j >= 0; j--)
a[i][j] = m++;
j++;
}
else
{
for (i++; j < lap; j++)
a[i][j] = m++;
j--;
for (i--; i >= 0; i--)
a[i][j] = m++;
i++;
}
lap++;
}
Print(n);
}
void Rectangle2()
{
int i = 0, j = 0;
int s = 1, nNum = 1;//s标记升降方向,斜向上为升s=1,斜向下为降s=-1
int n = 4;
while (1)
{
if (s == 1)
{
a[i][j] = nNum;
if (i - 1 < 0)
{
if (j + 1 == n)
i++;
else
j++;
s = -1;
}
else if (j + 1 == n)
{
i++;
s = -1;
}
else
{
i--;
j++;
}
}
else
{
a[i][j] = nNum;
if (j - 1 < 0)
{
if (j + 1 == n)
j++;
else
i++;
s = 1;
}
else if (i + 1 == n)
{
j++;
s = 1;
}
else
{
i++;
j--;
}
}
nNum++;
if (nNum > n*n)
break;
}
Print(n);
}
void Rectangle3()
{
int i = 0, j = 0;
int n = 4;
int x1=0, x2=n, y1=0, y2=n;//分别表示上下左右
int m = 1, s = 1;//s=1表示升序,s=-1表示降序
while (1)
{
if (s == 1)
{
for (j; j < y2; ++j)
a[i][j] = m++;
j--;
i++;
y2--;
for (i; i < x2; ++i)
a[i][j] = m++;
i--;
j--;
x2--;
s = -1;
}
else
{
for (j; j >=y1; --j)
a[i][j] = m++;
j++;
i--;
y1++;
for (i; i>=x1+1; --i)
a[i][j] = m++;
i++;
j++;
x1++;
s = 1;
}
if (m > n*n)
break;
}
Print(n);
}
void Rectangle4()
{
int i = 0, j = 0;
int n = 4;
int m = n*n;
int x1 = 0, y1 = 0, x2 = n, y2 = n;
int s = 1;
if (n % 2 == 0)
{
j = n - 1;
y2 = n - 1;
s = 1;
}
else
{
i = n - 1;
y1 = 1;
s = -1;
}
while (1)
{
if (s == 1)
{
for (i; i < x2; ++i)
a[i][j] = m--;
i--;
j--;
x2--;
for (j; j >= y1; --j)
a[i][j] = m--;
j++;
i--;
y1++;
s = -1;
}
else
{
for (i; i >=x1; --i)
a[i][j] = m--;
i++;
j++;
x1++;
for (j; j
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