10161 - 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

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

#include<stdio.h>

#include<math.h>

int main(void)

{

	int n;

	while(scanf("%d",&n)&&n)

	{

		int x=0,y=0;

		int p=sqrt((double)n);

		if(p*p==n&&p%2) {x=1;y=p;}

		else if(p*p==n&&p%2==0){x=p;y=1;}

		else if(p%2==0&&n!=p*p) {

			if(n-p-1<=p*p){x=p+1;y=n-p*p;}

			else{x=p-(n-1-p-p*p)+1;y=p+1;}

		}

		else if(p%2==1&&n!=p*p) {

			if(n-p-1<=p*p){x=n-p*p;y=p+1;}

			else{x=p+1;y=p-(n-1-p-p*p)+1;}

		}

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

	}

	return 0;

}

巴特西