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 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。

 1 #include <iostream>
 2 #include <iomanip>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <cmath>
 6 #include <algorithm>
 7 #include <cstdio>
 8 
 9 using namespace std;
10 
11 int main(void)
12 {
13     int n;
14     while (cin >> n)
15     {
16         if(!n)    break;
17         int k = floor(sqrt(n));
18         if (k * k == n)    { if(k&1)cout<<"1 "<<k<<endl;else cout<<k<<" 1"<<endl; }
19         else
20         {
21             int t = n - k * k;
22             if (k&1)
23             {
24                 if (t <= k + 1)    cout<<n-k*k<<' '<<k+1<<endl;
25                 else    cout<<' '<<k+1<<k+1-((n-k*k)%(k+1))<<endl;
26             }
27             else
28             {
29                 if (t<=k+1)    cout<<k+1<<' '<<n-k*k<<endl;
30                 else    cout<<k+1-((n-k*k)%(k+1))<<' '<<k+1<<endl;
31             }
32         }
33     }
34 
35     return 0;
36 }
posted on 2012-11-08 21:13  aries__liu  阅读(321)  评论(0编辑  收藏  举报