一道数学水题，找找规律。
首先要判断给的数在第几层，比如说在第n层。然后判断(n * n - n + 1)(其坐标也就是(n,n)) 之间的关系。
还要注意n的奇偶。
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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 |
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
AC代码：
//#define LOCAL
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std; int main(void)
{
#ifdef LOCAL
freopen("10161in.txt", "r", stdin);
#endif int N;
while(scanf("%d", &N) == && N)
{
int n = (int)ceil(sqrt(N));
int x, y;
if(n & == )
{
if(N < n * n - n + )
{
x = n;
y = N - (n - ) * (n - );
}
else
{
y = n;
x = n * n - N + ;
}
}
else
{
if(N < n * n - n + )
{
y = n;
x = N - (n - ) * (n - );
}
else
{
x = n;
y = n * n - N + ;
}
} cout << x << " " << y << endl;
}
return ;
}
代码君