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) == 1 && N)
    {
        int n = (int)ceil(sqrt(N));
        int x, y;
        if(n & 1 == 1)
        {
            if(N < n * n - n + 1)
            {
                x = n;
                y = N - (n - 1) * (n - 1);
            }
            else
            {
                y = n;
                x = n * n - N + 1;
            }
        }
        else
        {
            if(N < n * n - n + 1)
            {
                y = n;
                x = N - (n - 1) * (n - 1);
            }
            else
            {
                x = n;
                y = n * n - N + 1;
            }
        }

        cout << x << " " << y << endl;
    }
    return 0;
}