uva 10161 Ant on a Chessboard

uva 10161 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 8th second , she was at (2,3), and at 20th 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
题目大意：找到数字对应的行和列。
解题思路：找规律，奇数行，起始为行数的平方。偶数列，起始为列数的平方。行和列有与数字匹配的规律。

#include<stdio.h>
int main(){

		int N, n, cnt, x, y;
	while (scanf("%d", &N) != EOF && N) {//用笨办法做的，模拟全过程
		 n = 0;							//应该先找规律，总结，然后按公式来写代码
		 cnt = 1;
		 x = 1;
		 y = 1;
		for (int i = 1; i < N; ) {
			y++;
			i++;
			if (i==N) {
				break;	
			}
				x+=cnt;
				i+=cnt;
				if (i >= N) {
					x-=cnt;
					i-=cnt;
					for (int j = 0; j < cnt; j++){
						x++;
						i++;
						if (i==N) {
							break;	
						}
					}
				}
			if (i==N) {
				break;	
			}
			y-=cnt;
			i+=cnt;
			if (i >= N) {
					y+=cnt;
					i-=cnt;
					for (int j = 0; j < cnt; j++){
						y--;
						i++;
						if (i==N) {
							break;	
						}
					}
				}
				if (i==N) {
				break;	
			}
			x++;
			i++;
			if (i==N) {
				break;	
			}
			cnt++;
			y+=cnt;
			i+=cnt;
			if (i >= N) {
					y-=cnt;
					i-=cnt;
					for (int j = 0; j < cnt; j++){
						y++;
						i++;
						if (i==N) {
							break;	
						}
					}
				}
			if (i==N) {
				break;	
			}
			x-=cnt;
			i+=cnt;
			if (i >= N) {
					x+=cnt;
					i-=cnt;
					for (int j = 0; j < cnt; j++){
						x--;
						i++;
						if (i==N) {
							break;	
						}
					}
				}
			if (i==N) {
				break;
			}
			cnt++;
		}	
		printf("%d %d\n", x, y);
	}
	return 0;
}