UVa 10161 - Ant on a Chessboard

最新推荐文章于 2017-09-06 14:51:51 发布

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本文介绍了一种算法，用于确定一只蚂蚁按照特定蛇形路径走过M*M棋盘上每个格子的位置。通过分析蚂蚁行走的规律，文章提供了一个C语言实现的例子，能够计算在任意给定时间内蚂蚁所处的具体坐标。

Problem A.Ant on a Chessboard

Background

One day, an ant called Alice came to an M*M chessboard. She wanted to go around all the grids. So she began to walk along the chessboard according to this way: (you can assume that her speed is one grid per second)

At the first second, Alice was standing at (1,1). Firstly she went up for a grid, then a grid to the right, a grid downward. After that, she went a grid to the right, then two grids upward, and then two grids to the left…in a word, the path was like a snake.

For example, her first 25 seconds went like this:

( the numbers in the grids stands for the time when she went into the grids)

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

1 2 3 4 5

At the 8^th second , she was at (2,3), and at 20^th second, she was at (5,4).

Your task is to decide where she was at a given time.

(you can assume that M is large enough)

Input

Input file will contain several lines, and each line contains a number N(1<=N<=2*10^9), which stands for the time. The file will be ended with a line that contains a number 0.

Output

For each input situation you should print a line with two numbers (x, y), the column and the row number, there must be only a space between them.

Sample Input

Sample Output

2 3

5 4

1 5

此题明白了过程，程序就会写出：

#include <stdio.h>
#include <math.h>
int main()
{
    long long int  n,s,t,X,Y;
    while(scanf("%lld",&t)!=EOF)
    {
        if(t==0)
        {
            break;
        }
        if(t<4)
        {
            if(t==1)
            {
                X=1;Y=1;
            }else if(t==2)
            {
                X=2;Y=1;
            }else if(t==3)
            {
                X=2;Y=2;
            }
        }else
        {
            n=2;
            while(n*n<t)
            {
                n+=2;
            }
            if(n*n==t)
            {
                X=1;Y=n;
            }else
            {
                n-=2;
                X=1;Y=n+1;
                s=t-n*n;
                n+=1;
                if(s<=(2*n-1))
                {
                    if(s<=n)
                    {
                        X=s;
                    }else
                    {
                        X=n;
                        s-=n;
                        Y-=s;
                    }
                }else
                {
                    X=n;Y=1;
                    s-=2*n-1;
                    n+=1;
                    if(s<=n)
                    {
                        X+=1;
                        Y=s;
                    }else
                    {
                        X+=1;
                        s-=n;
                        Y=n;
                        X-=s;
                    }
                }
            }
        }
        printf("%lld %lld\n",Y,X);
    }
    return 0;
}