编程练习-螺旋数组

题目如下 =========== 43 44 45 46 47 .. ..
42 21 22 23 24 25 26
41 20 07 08 09 10 27
40 19 06 01 02 11 28
39 18 05 04 03 12 29
38 17 16 15 14 13 30
37 36 35 34 33 32 31 看清以上数字排列的规律, 设1点的坐标是(0,0), x方向向右为正, y方向向下为正. 例如, 7的坐标为(-1，-1), 2的坐标为(0,1), 3的坐标为(1,1). 编程实现输入任意点的坐标, 输出所对应的数字. =========== 我自己做的答案贴在下面. 在VC上调试通过.

标准答案还没看, 太晚了. 改天再研究一下贴上来吧.

#include <iostream>
#include <cmath>

using namespace std;

int iMatrix[100][100];

void FillEdge(int width, int layer)
{
// Up right corner
int temp = iMatrix[layer][width - 1 - layer] = (width - layer - layer) * (width - layer -layer);

// Up edge
for( int i = width - 1 - layer; i >= layer; --i)
iMatrix[layer][i] = temp--;

// Left edge
for( int j = 1 + layer; j < width - layer; ++j)
iMatrix[j][layer] = temp--;

// Low edge
for( int i = 1 + layer; i < width - layer; ++i)
iMatrix[width - 1 - layer][i] = temp--;

// Right edge
for( int j = width - 2 - layer; j > layer; --j)
iMatrix[j][width - 1 - layer] = temp--;
}

void Print(int width)
{
for(int i = 0; i < width; i++)
{
for(int j = 0; j < width; j++)
{
cout<<iMatrix[i][j]<<"\t";
}
cout<<endl;
}
cout<<endl;
}

int main()
{
memset(iMatrix, 0, sizeof(iMatrix));

cout << "螺旋数组" << endl;

int x = 0;
int y = 0;

cin >> x;
cin >> y;

int iBigger = 0;
if(abs(x) >= abs(y))
iBigger = abs(x);
else
iBigger = abs(y);

int width = 2 * iBigger + 1;

for(int w = 0; w <= iBigger; w++)
{
FillEdge(width, w);
}

Print(width);

cout<<iMatrix[y + iBigger][x + iBigger];
}

题目来源:

<程序员面试宝典>