UVa 10161 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)

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