UVA 10161 --- Problem A.Ant on a Chessboard 找规律

Problem A.Ant on a Chessboard

  Time Limit:3000MS     Memory
Limit:0KB     64bit IO Format:%lld
& %llu

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。。。。

#include <stdio.h>
#include <math.h>
#include <string.h>

int main()
{
int k,a,b,x,y,c,d,f;
while(scanf("%d",&k),k)
{
a = int(sqrt(k * 1.0));
b = a * a;
if(b == k)
{
if(a & 1)
{
x = 1;
y = a;
}
else
{
x = a;
y = 1;
}
}
else
{
c = (a + 1) * (a + 1);
d = (c - b) >> 1;
f = b + d + 1;
if(a & 1)
{
if(k >= f)
{
x = a + 1;
y = a + 1 - k + f;
}
else
{
x = a + 1 - f + k;
y = a + 1;
}
}
else
{
if(k >= f)
{
x = a + 1 -k + f;
y = a + 1;
}
else
{
x = a + 1;
y = a + 1 - f + k;
}
}
}
printf("%d %d\n",x,y);
}

return 0;
}