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 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;}