Problem A.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 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

找规律啊，使劲的找啊~

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