UVa 10161 Ant on a Chessboard

一道数学水题，找找规律。

首先要判断给的数在第几层，比如说在第n层。然后判断(n * n - n + 1)(其坐标也就是(n,n)) 之间的关系。

还要注意n的奇偶。

 

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 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

 

AC代码：

 1 //#define LOCAL

 2 #include <iostream>

 3 #include <cstdio>

 4 #include <cstring>

 5 #include <cmath>

 6 using namespace std;

 7 

 8 int main(void)

 9 {

10     #ifdef LOCAL

11         freopen("10161in.txt", "r", stdin);

12     #endif

13 

14     int N;

15     while(scanf("%d", &N) == 1 && N)

16     {

17         int n = (int)ceil(sqrt(N));

18         int x, y;

19         if(n & 1 == 1)

20         {

21             if(N < n * n - n + 1)

22             {

23                 x = n;

24                 y = N - (n - 1) * (n - 1);

25             }

26             else

27             {

28                 y = n;

29                 x = n * n - N + 1;

30             }

31         }

32         else

33         {

34             if(N < n * n - n + 1)

35             {

36                 y = n;

37                 x = N - (n - 1) * (n - 1);

38             }

39             else

40             {

41                 x = n;

42                 y = n * n - N + 1;

43             }

44         }

45 

46         cout << x << " " << y << endl;

47     }

48     return 0;

49 }

代码君