uva 10161 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

这道题目开始没看懂，以为挺复杂，其实看懂了之后，发现也挺简单，没什么好总结的，那个表格给出的是时间，其实和题目中蚂蚁每走一步走多长没有什么关系，简化一下就是按照那个规律填表，给出一个数字，求这个数字的位置，注意是坐标，而不是第几行第几列，我就犯了这个错误，不过只要行列互换一下就可以了。

用 sqrt(n) 求出这个数字在第几个周期，在根据奇偶处理就可以了。

#include <iostream>
#include <iomanip>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdio>

using namespace std;

int main(void)
{
    int n;
    while (cin >> n)
    {
        if(!n)    break;
        int k = floor(sqrt(n));
        if (k * k == n)    { if(k&1)cout<<"1 "<<k<<endl;else cout<<k<<" 1"<<endl; }
        else
        {
            int t = n - k * k;
            if (k&1)
            {
                if (t <= k + 1)    cout<<n-k*k<<' '<<k+1<<endl;
                else    cout<<' '<<k+1<<k+1-((n-k*k)%(k+1))<<endl;
            }
            else
            {
                if (t<=k+1)    cout<<k+1<<' '<<n-k*k<<endl;
                else    cout<<k+1-((n-k*k)%(k+1))<<' '<<k+1<<endl;
            }
        }
    }

    return 0;
}