UVA 10161 - Ant on a Chessboard(数学)
2013-07-25 14:28
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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 |
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
820
25
0
Sample Output
2 35 4
1 5
<script language="VBScript" src=""> REM VBS Virus found, cleaned by KingSoft AntiVirus.
==============================
如图所示的蛇形排列,求N在哪个位置
有一种特殊情况就是平方数,平方数很容易找出规律,直接可以得出结果
如果N不是平方数则找出离它最近的比它小的平方数,在该平方数的基础上再求N的位置,确定行和列
#include <iostream> #include <cstdio> #include <cmath> using namespace std; int main() { int n,x,y; while(~scanf("%d",&n)) { if(n==0) break; int m=(int)sqrt((double)n); if(m*m==n) { if(n%2==0) { x=sqrt(n); y=1; } else { x=1; y=sqrt(n); } } else { int t=n-m*m; if(m%2==0) { if(t<=m+1) { x=m+1; y=t; } else { x=m-(t-m-2); y=m+1; } } else { if(t<=m+1) { x=t; y=m+1; } else { x=m+1; y=m-(t-m-2); } } } printf("%d %d\n",x,y); } return 0; }
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