UVa 10161 Ant on a Chessboard (简单数学)

最新推荐文章于 2019-02-19 11:45:02 发布

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本文探讨了AI在游戏开发领域的应用，包括游戏程序、游戏美术、游戏策划等多个方面，展示了AI如何提升游戏体验和效率。

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

。.....A.C.............................................................................................................................................................................

#include<stdio.h>

#include<math.h>
int main()
{
int step;
while (scanf("%d",&step)!=EOF,step>=1)
{
int column = (int)ceil(sqrt(step));
int diagonal = column * (column - 1) + 1;
if (column&1)//判断奇偶性
{
if (step >= diagonal)
printf("%d %d\n",column - (step - diagonal),column);
else
printf("%d %d\n",column,column - (diagonal - step));
}
else//奇数
{
if (step >= diagonal)
printf("%d %d\n",column,column-(step-diagonal));
else
printf("%d %d\n",column-(diagonal-step),column);
}
}
return 0;
}