UVA 10161 Ant on a Chessboard（规律）

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 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

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

这道题是让你根据题目中给出的规律，来判断给的数字n是几列几行。首先，我们可以先根据n的值来判断M最小为多少(n的行或列总有一个是M)，然后我们可以发现对角线上的值是有规律的 = i*(i-1)+1，当M为偶数的时候，对角线上的值往左递减，往下递加，M为积数的时候正好相反，根据它们之间的值来判断另一个行或列。

#include <iostream>
using namespace std;
int main()
{
  int n;
  while(cin>>n)
  {
      if(!n)
        break;
     int k=n;
     for(int i=1;i<n;i++)
     {
         if(i*i==n)
         {
             k=i;
             break;
         }
         else if(n>i*i && n<((i+1)*(i+1)))
         {
             k=i+1;
             break;
         }
     }
    int s=k*(k-1)+1;
    if(k%2==1)
    {
        if(n>=s)
         cout<<k-(n-s)<<" "<<k<<endl;
        else cout<<k<<" "<<k-(s-n)<<endl;

    }
    else
    {
        if(n>=s)
         cout<<k<<" "<<k-(n-s)<<endl;
         else cout<<k-(s-n)<<" "<<k<<endl;
    }
  }
  return 0;
}