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
找规律:副对角线之间差值成等差数列,且根据给出的n可求出所在的最小方格数m,于是可求出对角线的值=(m-1)^2-m,在根据m的奇偶性,左、下的变化规律即可求得。
#include <iostream> #include <cmath> using namespace std; int main() { int n,r,c,m,dia; while(cin>>n&&n) { m=sqrt((double)n); if(m*m<n) m++; r=c=m; dia=(m-1)*(m-1)+m; if(m%2==0) { if(dia>=n) c-=dia-n; else r-=n-dia; } else { if(dia>=n) r-=dia-n; else c-=n-dia; } cout<<c<<' '<<r<<endl; } return 0; }