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：

 //#define LOCAL
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 using namespace std;

 int main(void)
 {
     #ifdef LOCAL
         freopen("10161in.txt", "r", stdin);
     #endif

     int N;
     while(scanf("%d", &N) ==  && N)
     {
         int n = (int)ceil(sqrt(N));
         int x, y;
         if(n &  == )
         {
             if(N < n * n - n + )
             {
                 x = n;
                 y = N - (n - ) * (n - );
             }
             else
             {
                 y = n;
                 x = n * n - N + ;
             }
         }
         else
         {
             if(N < n * n - n + )
             {
                 y = n;
                 x = N - (n - ) * (n - );
             }
             else
             {
                 x = n;
                 y = n * n - N + ;
             }
         }

         cout << x << " " << y << endl;
     }
     return ;
 }

代码君