uva 10161 - Ant on a Chessboard
2012-11-08 21:13
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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 |
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
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。
#include <iostream> #include <iomanip> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #include <cstdio> using namespace std; int main(void) { int n; while (cin >> n) { if(!n) break; int k = floor(sqrt(n)); if (k * k == n) { if(k&1)cout<<"1 "<<k<<endl;else cout<<k<<" 1"<<endl; } else { int t = n - k * k; if (k&1) { if (t <= k + 1) cout<<n-k*k<<' '<<k+1<<endl; else cout<<' '<<k+1<<k+1-((n-k*k)%(k+1))<<endl; } else { if (t<=k+1) cout<<k+1<<' '<<n-k*k<<endl; else cout<<k+1-((n-k*k)%(k+1))<<' '<<k+1<<endl; } } } return 0; }
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