Prairie dogs - ecbtnrt的博客 - 编程中国

首页新闻小组威客人才下载博客代码贴在线编程论坛

注册 | 登录

Prairie dogs

作者在 2007-06-01 18:08:00 发布以下内容

Oh, my God! The lovely prairie dogs come again! We know they are very naughty and always play some funny games. This time, they play a game named Spiral Queue.

Each of the prairie dogs has a number and they stand in a funny queue named Spiral Queue like in Figure 1.

Given the coordinate, the direction of x-axis and y-axis is indicated in Figure 2. We suppose the coordinate of 1 is (0,0), then the coordinate of 2 is (1,0), the coordinate of 3 is (1,1) and the coordinate of 7 is (-1,-1).

21   22  ...
20   7   8   9  10
19   6   1   2  11
18   5   4   3  12
17  16  15  14  13
      
      Figure1

Find out the laws of the Spiral Queue.

Your task is here: Given x and y (-1000 ≤ x,y ≤ 1000), the coordinate of a prairie dog, try to find out the number of the prairie dog.

Input

Each line of input will have two numbers x and y, indicating the coordinate of a prairie dog. The input stops when EOF (end of file) is reached.

Output

You must output the number of the prairie dog based on the coordinate (x,y), followed by a newline.

Sample Input

0 0 -1 -1 1 0 1 1 100 100 1000 -1000

Sample Output

1 7 2 3 39801 4004001

经典程序 | 阅读 2805 次

收藏

文章评论，共3条

ecbtnrt(作者)

2007-06-01 18:12

1

#include        <stdio.h>

int x_max(int m ,int n)
{
    if(m < 0)
        m = -m;
    if(n < 0)
        n = -n;
    if(m > n)
        return m;
    return n;
}

int main(void)
{
    int x, y;
    int sum = 0;
    int t;
    int i;

    scanf("%d%d", &x, &y);
    t = x_max(x, y);
    sum = 4 * (t -1)*(t - 1) + 4 * (t - 1) + 2 ;

    for( i = 0; i <= (2*t-1); i++)
    {
        if(y == -t+1+i)
        {
            if(x == t)
            {
                sum += i;
                printf("%d\n", sum);
                return 0;
            }
        }
    }
    sum += 2*t-1;

    for(i = 1; i <= 2*t; i++)
    {
        if(x == t - i)
        {
            if(y == t)
            {
                sum += i;
                printf("%d\n", sum);
                return 0;
            }
        }
    }
    sum += 2*t;

    for(i = 1; i <= 2*t; i++)
    {
        if(y == t - i)
        {
            if(x == -t)
            {
                sum += i;
                printf("%d\n", sum);
                return 0;
            }
        }
    }
    sum += 2*t;

    for(i = 0; i <= 2*t; i++)
    {
        if(x == -t+i)
        {
            sum += i;
            printf("%d\n", sum);
            return 0;
        }
    }
    printf("%d\n", sum );
    return 0;

}

赞回复

ecbtnrt(作者)

2007-06-01 18:13

2

这是我写的程序，运行是正确的。不过如果提交的话有点问题，我搞不懂为什么。你能指导我一下吗

赞回复

ecbtnrt(作者)

2007-06-04 08:34

3

我想我找到原因了，是不是要把输入输出改成文件得形式

赞回复

ecbtnrt的博客

ecbtnrt

浏览107743次

文章分类

文章归档

最新评论