UVA10161 - Ant on a Chessboard

最新推荐文章于 2019-02-19 11:45:02 发布

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本文介绍了一道经典的路径寻找问题——蚂蚁在棋盘上的行走路径。通过算法解析，我们能够确定在任意给定的时间点上，蚂蚁所处的具体位置。文章提供了完整的C语言实现代码。

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题目：

Problem A.Ant on a Chessboard

Background

One day, an ant calledAlice came to an M*M chessboard. She wanted to go around all the grids. So shebegan to walk along the chessboard according to this way: (you can assume thather 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 theright, a grid downward. After that, she went a grid to the right, then twogrids upward, and then two grids to the left…in a word, the path was like asnake.

For example, her first25 seconds went like this:

( the numbers in thegrids 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^thsecond, 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 willcontain 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 anumber 0.

Output

For each input situationyou 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

解答：

#include<stdio.h>
#include<stdlib.h>
int main()
{
	long n;
	while(scanf("%ld",&n)!=EOF)
	{
		if(n==0)
			break;
		long s=0,i=1;
		for(;;)
		{
			s+=i;
			if(s>=n)
			{
				if(s%2==1)
				{
					if(s-n>i/2)
						printf("%ld %ld\n",i/2+1,i+n-s);
					else
						printf("%ld %ld\n",s+1-n,i/2+1);
					break;
				}
				else
				{
					if(s-n>i/2)
						printf("%ld %ld\n",i+n-s,i/2+1);
					else
						printf("%ld %ld\n",i/2+1,s+1-n);
					break;
				}
			}
			i+=2;
		}
	}
	return 0;
}