Ultraviolet A 10161 ant on a chessboard

A mathematical question, look for patterns.

First, you must determine the number at the nth layer, for example, at the nth layer. Then, judge the relationship between (N * n-n + 1) (its coordinates are (n, n.

Pay attention to the parity of N.

 

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 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)

 

 

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 shocould 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 code:

 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 }

Code Jun