UVa 10161 Ant on a Chessboard

最新推荐文章于 2019-02-19 11:45:02 发布

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最新推荐文章于 2019-02-19 11:45:02 发布

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分类专栏： UVa 文章标签： UVa 刘汝佳数学

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本文介绍了一只蚂蚁在一个M*M的棋盘上行走的路径规律。蚂蚁从左上角开始，按照特定的蛇形路径遍历整个棋盘。文章详细解释了如何根据给定的时间确定蚂蚁所在的具体位置，并提供了一段AC代码实现。

一道数学水题，找找规律。

首先要判断给的数在第几层，比如说在第n层。然后判断(n * n - n + 1)(其坐标也就是(n,n)) 之间的关系。

还要注意n的奇偶。

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

AC代码：

//#define LOCAL
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

int main(void)
{
	#ifdef LOCAL
		freopen("10161in.txt", "r", stdin);
	#endif

	int N;
	while(scanf("%d", &N) == 1 && N)
	{
		int n = (int)ceil(sqrt(N));
		int x, y;
		if(n & 1 == 1)
		{
			if(N < n * n - n + 1)
			{
				x = n;
				y = N - (n - 1) * (n - 1);
			}
			else
			{
				y = n;
				x = n * n - N + 1;
			}
		}
		else
		{
			if(N < n * n - n + 1)
			{
				y = n;
				x = N - (n - 1) * (n - 1);
			}
			else
			{
				x = n;
				y = n * n - N + 1;
			}
		}

		cout << x << " " << y << endl;
	}
	return 0;
}