最近无聊的很,看了看一些基础的面试题,虽说简单点,但还是能学到一点东西的。
就比方main函数结束之后还能执行函数吗?答案是可以,使用atexit注册一个函数就可以了。
下面重点谈谈这个螺旋矩阵!
关于螺旋矩阵的说法不一,这里指的是形如
21 22................
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13
的矩阵。
问题有两个:
1. 编程实现输出这个矩阵
2. 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).编程实现输入任意一点坐标(x,y),输出所对应的数字。
1. 第一个问题我是采用模拟进行构造的,可以看到从1开始的方向变化始终是 right->down->left->up,
所持续走的长度为1->1->2->2->3->3->...,发现了这个规律不难写出代码了!注意下面我把1的位置设置
在((n-1)/2, (n-1)/2)的位置。
void Simulate(int n)
![](https://www.cnblogs.com/Images/OutliningIndicators/ExpandedBlockStart.gif)
![](https://www.cnblogs.com/Images/OutliningIndicators/ContractedBlock.gif)
{
int x, y;
x = y = (n - 1) / 2; //1的位置
data[x][y] = 1;
int len = 1;
int count = 0;
int num = 2;
DIRECTION dir = RIGHT;
while(num <= n * n)
![](https://www.cnblogs.com/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
{
for(int i = 0; i < len; i++)
![](https://www.cnblogs.com/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
{
switch(dir)
![](https://www.cnblogs.com/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
{
case LEFT:
--y; break;
case RIGHT:
++y; break;
case UP:
--x; break;
case DOWN:
++x; break;
default: break;
}
data[x][y] = num++;
}
count++;
if(count == 2)
![](https://www.cnblogs.com/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
{
count = 0;
len++;
}
dir = (DIRECTION)((dir + 1) % 4);
}
}
2. 第二个问题我也是先找出规律,然后进行模拟。
首先,不难看出n*n的螺旋矩阵的右下角的坐标一定是(m, m),这里m=n-1
通过观察,可以看出 n=1的时候,右下角(0,0)的值为1,当n=2的时候,右下角(1,1)的坐标值为(3,3),当n=3的时候,右下角(2,2)的坐标值为13.直觉告诉我,这个值是关于n的二次函数,设f(n) = a*n^2 + b*n + c
联立方程组,可以求得a,b,c。 最终算出来的f(n) = 4*n^2 - 2*n + 1
下面再根据(x,y)和右下角(n-1,n-1)之间的关系,计算出值即可。这里要注意当x的值与n-1相同时,应优先考虑y与-m是否有联系。这就要求在函数中要注意x,y的判断先后顺序了。
代码如下:
//以(1,1)所在位置作为原点,向右作为x正半轴,向下作为y正半轴
int GetValue(int x, int y)
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![](https://www.cnblogs.com/Images/OutliningIndicators/ContractedBlock.gif)
{
int m = max(abs(x), abs(y));
int rightBottom = m * m * 4 - 2 * m + 1;
int value = 0;
if(x == -m)
![](https://www.cnblogs.com/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
{
value = rightBottom + 2 * m + m - y;
}
else if( y == m)
![](https://www.cnblogs.com/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
{
value = rightBottom + m - x;
}
else if(y == -m)
![](https://www.cnblogs.com/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
{
value = rightBottom + 4 * m + x + m;
}
else if( x == m )
![](https://www.cnblogs.com/Images/OutliningIndicators/ExpandedSubBlockStart.gif)
{
value = rightBottom - (m - y);
}
![](https://www.cnblogs.com/Images/OutliningIndicators/InBlock.gif)
return value;
}
![](https://www.cnblogs.com/Images/OutliningIndicators/None.gif)
完整代码如下:
![](https://www.cnblogs.com/Images/OutliningIndicators/ContractedBlock.gif)
![](https://www.cnblogs.com/Images/OutliningIndicators/ExpandedBlockStart.gif)
#include <iostream>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int N = 100;
int data[N + 1][N + 1];
enum DIRECTION
{
RIGHT, DOWN , LEFT, UP
};
//模拟整个过程
void Simulate(int n)
{
int x, y;
x = y = (n - 1) / 2; //1的位置
data[x][y] = 1;
int len = 1;
int count = 0;
int num = 2;
DIRECTION dir = RIGHT;
while(num <= n * n)
{
for(int i = 0; i < len; i++)
{
switch(dir)
{
case LEFT:
--y; break;
case RIGHT:
++y; break;
case UP:
--x; break;
case DOWN:
++x; break;
default: break;
}
data[x][y] = num++;
}
count++;
if(count == 2)
{
count = 0;
len++;
}
dir = (DIRECTION)((dir + 1) % 4);
}
}
//打印螺旋矩阵
void Output(int n)
{
int i, j;
for(i = 0; i < n; i++)
{
cout << data[i][0];
for(j = 1; j < n; j++)
cout << "\t" << data[i][j];
cout << endl;
}
}
//以(1,1)所在位置作为原点,向右作为x正半轴,向下作为y正半轴
int GetValue(int x, int y)
{
int m = max(abs(x), abs(y));
int rightBottom = m * m * 4 - 2 * m + 1;
int value = 0;
if(x == -m)
{
value = rightBottom + 2 * m + m - y;
}
else if( y == m)
{
value = rightBottom + m - x;
}
else if(y == -m)
{
value = rightBottom + 4 * m + x + m;
}
else if( x == m )
{
value = rightBottom - (m - y);
}
return value;
}
void TestPos(int n)
{
int i, j;
for(i = 0; i < n; i++)
{
cout << GetValue(0 - (n - 1) / 2, i - (n - 1) / 2);
for(j = 1; j < n; j++)
cout << "\t" << GetValue(j - (n - 1) / 2, i - (n - 1) / 2);
cout << endl;
}
}
int main()
{
int n;
while(cin >> n)
{
if(n <= 0 || n > 100)
{
cerr << "Size error!" << endl;
break;
}
else
{
Simulate(n);
Output(n);
cout << "*******************" << endl;
TestPos(n);
}
}
return 0;
}