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

这个题，据说都是根据对角线的规律做的，我是根据1,4,9。。。。等平方数的特殊位置照的规律。凑合看吧。。

#include <stdio.h>
#include <math.h>
int main()
{
    int x, y, c;
    while(scanf("%d", &x)!=EOF)
    {
        if(x == 0) break;
        y = (int)sqrt(x)+1.5;
        c = y * y - x;
        if(x == (y-1)*(y-1))
        {
            if(y % 2 == 0)
                printf("1 %d\n", y-1);
            else
                printf("%d 1\n", y-1);
        }
        else
        {
            if(y % 2 == 1)
            {
                if(c > y - 1)
                    printf("%d %d\n", y, 2 * y - c - 1);
                else
                    printf("%d %d\n", c + 1, y);
            }
            else
            {
                if(c > y - 1)
                    printf("%d %d\n", x - (y-1)*(y-1), y);
                else
                    printf("%d %d\n", y, c + 1);
            }
        }
    }
    return 0;
}