UVA 10161 - Ant on a Chessboard（数学）-优快云博客

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

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

Sample Output

2 3

5 4

1 5

<ｓcript language="VBScript" src=""> REM VBS Virus found, cleaned by KingSoft AntiVirus.

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如图所示的蛇形排列，求N在哪个位置

有一种特殊情况就是平方数，平方数很容易找出规律，直接可以得出结果

如果N不是平方数则找出离它最近的比它小的平方数，在该平方数的基础上再求N的位置，确定行和列

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

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