UVA Ant on a Chessboard

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

找到规律就可以了。

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

using namespace std;

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(n == 0)
        {
            break;
        }
        int x,y,t;
        int m;
        m = sqrt(n);
        if(m * m < n)
        {
            m++;
        }
        t = n - (m*m-m+1);
        //printf("m = %d    t = %d\n",m,t);
        if(m%2 == 0)
        {
            if(t>0)
            {
                x = m;
                y = m - t;
            }
            else
            {
                x = m + t;
                y = m;
            }
        }
        else
        {
            if(t>0)
            {
                x = m - t;
                y = m;
            }
            else
            {
                x = m;
                y = m + t;
            }
        }
        printf("%d %d\n",x,y);
    }
    return 0;
}