UVA10161 - Ant on a Chessboard

开学了，从现在开始吧～

题目：

 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

解答：

#include<stdio.h>
#include<stdlib.h>
int main()
{
    long n;
    while(scanf("%ld",&n)!=EOF)
    {
        if(n==0)
            break;
        long s=0,i=1;
        for(;;)
        {
            s+=i;
            if(s>=n)
            {
                if(s%2==1)
                {
                    if(s-n>i/2)
                        printf("%ld %ld\n",i/2+1,i+n-s);
                    else
                        printf("%ld %ld\n",s+1-n,i/2+1);
                    break;
                }
                else
                {
                    if(s-n>i/2)
                        printf("%ld %ld\n",i+n-s,i/2+1);
                    else
                        printf("%ld %ld\n",i/2+1,s+1-n);
                    break;
                }
            }
            i+=2;
        }
    }
    return 0;
}