UVA 10161 (暑假-数学-B - Ant on a Chessboard)

 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

5

4

3

2

1

 

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

8

20

25

0

 

 

Sample Output

2 3

5 4

1 5

#include <cstdio>
#include <cstring>
#include <cmath>

int main() { 
	int n; 

	while (scanf("%d", &n) && n) {
		int k = ceil(sqrt(n));
		int count;
		count = (k-1)*(k-1);
		
		if (k % 2) {
			for (int j = 1; j <= k; j++) {
				count++;
				if (count == n)
					printf("%d %d\n", k, j);
			}
			for (int i = 2; i <= k; i++) {
				count++;
				if (count == n)
					printf("%d %d\n", k - i + 1, k);
			}
		}
		else {
			for (int i = 1; i <= k; i++) {
				count++;
				if (count == n)
					printf("%d %d\n", i, k);
			}
			for (int j = 2; j <= k; j++) {
				count++;
				if (count == n)
					printf("%d %d\n", k, k - j + 1);
			}
		}
		
	}
	return 0;
}