旋转方阵解决方法

旋转方阵
Description  

打印出一个旋转方阵,见Sample   Output.

Input  

输入一个整数n(1   <=   n   <=   20),   n为方阵的行数。

Output  

输出一个大小为n*n的距阵

Sample   Input  


5

Sample   Output  


1       16     15     14     13
2       17     24     23     12
3       18     25     22     11
4       19     20     21     10
5       6       7       8       9

Hint  

输出控制：每个数字占4位，居左。  
cout   < <   setw(4)   < <   setiosflags(ios::left)   < <   a[i][j];


------解决方案--------------------
// tt.cpp : Defines the entry point for the console application.
//


#include "iostream "
#include "iomanip "

using namespace std;

void temperatures(const double Fahrenheit, double& celsius, double& kelvin) {

celsius = 1.8 * Fahrenheit + 32;
kelvin = celsius + 273.16;
}

void printf_rmatrix(int raw) {

if (raw < 0)
return;

int** arry = new int*[raw];
for(int i = 0; i < raw; ++i)
arry[i] = new int[raw];

for (int x = 0; x < raw; ++x)
for (int y = 0; y < raw; ++y)
arry[x][y] = 0;

int tick = 1;
int x = 0;
int y = 0;
int state = 0;

while (true) {

arry[x][y] = tick++;

if (state == 0) {
if (x + 1 < raw && arry[x + 1][y] == 0)
x++;
else if (y + 1 < raw && arry[x][y + 1] == 0) {
y++;
state = 1;
}
else
break;
}
else if (state == 1) {
if (y + 1 < raw && arry[x][y + 1] == 0)
y++;
else if (x - 1 > = 0 && arry[x - 1][y] == 0) {
x--;
state = 2;
}
else
break;
}
else if (state == 2) {
if (x - 1 > = 0 && arry[x - 1][y] == 0)
x--;
else if (y - 1 > = 0 && arry[x][y - 1] == 0) {
y--;
state = 3;
}
else
break;
}
else if (state == 3) {
if (y - 1 > = 0 && arry[x][y - 1] == 0)
y--;
else if (x + 1 < raw && arry[x + 1][y] == 0) {
x++;
state = 0;
}
else
break;

}
else
break;

}

for (int x = 0; x < raw; ++x)
{
for (int y = 0; y < raw; ++y)
cout < < setw(4) < < setiosflags(ios::left) < < arry[x][y];
cout < <endl;
}

for(int i = 0; i < raw; ++i)
delete[] arry[i];
delete[] arry;

}

void main()
{
printf_rmatrix(10);