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

一道很简单的数学问题，给定一个棋盘和蚂蚁在上面爬行的规律，每次询问时间求蚂蚁的位置

很容易由等差数列求和公式求出棋盘对角线上的数字从左下角到右上角为n^2+n+1,(n=0,1,2,……)

由此可以确定所询问的数字在棋盘上的范围，进而可以算出坐标

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

using namespace std;

int main()
{
    long long t;
    while(scanf("%d",&t)&&t)
    {
        long long x,y;
        long long n=(sqrt(4*t-3)-1)/2;
        long long border=(n+1)*(n+1);
        if(n%2)
            if(t>border)
            {
                x=n+2;
                y=t-border;
            }
            else
            {
                x=n+1;
                y=border-t+1;
            }
        else
            if(t>border)
            {
                y=n+2;
                x=t-border;
            }
            else
            {
                y=n+1;
                x=border-t+1;
            }
        printf("%lld %lld
",x,y);
    }

    return 0;
}