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)   5 4 3 2 1   1          2          3           4           5 At the 8th second , she was at (2,3), and at 20th 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 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.       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.     8 20 25 0     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;
}