UVa10161 Ant on a Chessboard

最新推荐文章于 2020-02-07 11:05:28 发布

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最新推荐文章于 2020-02-07 11:05:28 发布

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本文介绍了一种算法，用于确定一只蚂蚁在一个正方形棋盘上行走的路径，并找到特定时刻蚂蚁的位置。通过数学计算，可以高效地求解任意给定时间点蚂蚁所在的格子坐标。

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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

这题其实就是给定一个数，要求找出这个数在矩阵中的位置。我是先将给定的数减1，保证第N层的数字都小与N*N且大于等于(N-1)*(N-1)。然后计算该数字与N*N及(N-1)*(N-1)的距离，取较小的，然后根据其所在层数的奇偶性，输出，具体可见代码。

#include <iostream>
#include <cmath>
using namespace std;

int main() {
	int num;
	int order;
	int p;
	int q;
	while (cin >> num) {
		if (num == 0)
			break;
		num--;
		order = (int)pow(num,0.5);
		p = num - order * order + 1;
		q = (order + 1) * (order + 1) - num;
		if (order % 2)
			if (p < q)
				cout << p << " " << order+1 << endl;
			else
				cout << order+1 << " " << q << endl;
		else
			if (p < q)
				cout << order+1 << " " << p << endl;
			else
				cout << q << " " << order+1 << endl;
	}
	return 0;
}