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 LOCAL11         freopen("10161in.txt", "r", stdin);12     #endif13 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             else27             {28                 y = n;29                 x = n * n - N + 1;30             }31         }32         else33         {34             if(N < n * n - n + 1)35             {36                 y = n;37                 x = N - (n - 1) * (n - 1);38             }39             else40             {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