数学B - Ant on a Chessboard

最新推荐文章于 2019-02-19 11:45:02 发布

原创最新推荐文章于 2019-02-19 11:45:02 发布 · 303 阅读

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本文介绍了一种在M*M棋盘上进行蛇形遍历的算法实现，通过跟踪一个角色Alice的移动路径，确定她在任意给定时间的位置。文章详细解析了蛇形遍历的规律，并提供了一个C语言实现的示例程序。

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 rst 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 leftin a word, the path was like a snake.

For example, her rst 25 seconds went like this:

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

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

is large enough).

Input

Input le will contain several lines, and each line contains a number

9

), which stands

for the time. The le will be ended with a line that contains a number `

0

Output

For each input situation you should print a line with two numbers (

), the column and the row

number, there must be only a space between them.

SampleInput

8

20

25

0

SampleOutput

2 3

5 4

1 5

这题太大不能写一个数组出来

首先想到要找规律

先去看平方数

去看了题解

要从对角线开始看

1 3 7 13 每两个数差2, 4, 6 数列首项1

num+= 2×（i - 1 ）

输入的数要和num比较大小比num大的又要《=num + i -1

比num小的又要 = num - i + 1

在判断i的奇偶来决定x变还是y变

不知道为什么1 2 有问题，偷懒把它们提出来

#include <stdio.h>

int main(){
	int n, num = 1, x, y;
	while (scanf("%d", &n) != EOF) {
		if (n == 0)
			break;
		num = 1;
		if (n == 1) {
			x = 1;
			y = 1;
		}
		else if (n == 2) {
			x = 1;
			y = 2;
		}
		else 
		for (int i = 1; i < n; i++) {
			num += 2 * (i - 1);
			if (num == n) {
				x = i;
				y = i;
				break;
			}

			if (n > num && n <= num + i - 1) {
				if (i % 2 == 0) {
					x = i;
					y = i - n + num;
				}
				else {
					y = i;
					x = i - n + num;
				}
				break;		
			}
			else if (n < num && n >= num - i + 1) {
				if (i % 2 == 0) {
					y = i;
					x = i - num + n;
				}
				else {
					x = i;
					y = i - num + n;
				}
				break;
			}

		}
		printf("%d %d\n", x, y);
	}
	return 0;
}