第三次训练——Problem A.Ant on a Chessboard

Problem A.Ant on a Chessboard 
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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

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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)

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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.

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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.

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Sample Input

8

20

25

0

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Sample Output

2 3

5 4

1 5

 

提示：观察对角线，找规律。

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

int main()
{
int m,cha,kf,pf,ysj;
while(scanf("%d",&m)!=EOF)
{
if(m != 0)
{
kf = sqrt(m);
if(kf*kf!=m)
{
pf = (kf + 1) * (kf + 1);
ysj = pf - kf;
cha = abs(ysj - m);
if(kf % 2 == 1)
{
if(ysj > m)
{
printf("%d %d\n",kf + 1 - cha,kf + 1);
}
else if(ysj < m)
{
printf("%d %d\n",kf + 1,kf + 1 -cha);
}
else
{
printf("%d %d\n",kf + 1,kf + 1);
}
}
else if(kf % 2 == 0)
{
if(ysj > m)
{
printf("%d %d\n",kf + 1,kf + 1 -cha);
}
else if(ysj < m)
{

printf("%d %d\n",kf + 1 - cha,kf + 1);
}
else
{
printf("%d %d\n",kf + 1,kf + 1);
}
}
}
else if(kf * kf == m)
{
if(kf % 2 == 1)
printf("1 %d\n",kf);
else if(kf % 2 == 0)
printf("%d 1\n",kf);
}
}
else
break;
}
return 0;
}