UVA-10161 - Ant on a Chessboard

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

 

题目要求给出给出N，求蚂蚁第N次爬到哪里，这个跟白书上的Canton数表题是如出一辄。。。找到数学规律，把它划分成8项加16项加24项。。即8K的一个等比数列，首先求出它在哪个区间，再从区间里按规则找出来。。表示做完后看了其他大神的题解，我表示自己找出的规律真的好挫。。。还写了2个小时才A掉。。。不过是1次A掉还是蛮爽的

还有就是要注意细节。。。殷犇说的，注意细节。。

 

 

#include <iostream>

using namespace std;

int main()
{
    int n;
    while (cin>>n&&n)
    {
      int s=1,k=1;
      int x,y;
      while (s<n)
      {
        s+=8*k;
        k++;
      }
      k--;
      int dis=s-n;
      int rec=dis/(4*k+1);
      if (rec)
      {
          if((dis-4*k)/(2*k))
          {
            x=2*k-(dis-6*k)%(2*k);
            y=2*k;
          }
          else
          {
              x=2*k;
              y=(dis-4*k)%(2*k);
          }
      }
       else
       {
           if (dis/(2*k+1))
           {
               x=2*k+1;
               y=2*k+1-(dis-2*k)%(2*k+1);
           }
           else
           {
               y=2*k+1;
               x=dis+1;
           }
       }

      cout<<x<<" "<<y<<endl;

    }

    return 0;
}