UVA10161 Ant on a Chessboard

这道题找规律 ，如果暴力基本超时了 ，找到规律就好做了 我的方法是把表格看成一层一层的 倒着的L型是一层 然后把层中间的数计算出来 与用户输入的数作差 再判断下奇偶，分情况讨论 算法复杂度虽然说是 O(1)但是 代码感觉还是繁琐了 有空再想想更简单的解法。

笔记：

ceil是向上取整

floor是向下取整

题目及AC代码如下

 

 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 8^th second , she was at (2,3), and at 20^th 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

#include<stdio.h>
#include<math.h>

int main(void)
{
    int n;
    while (scanf("%d", &n) && n!=0)
    {
        int l = ceil((sqrt(1.0*n)));            //l指的是这个数所处的层数
        int lmid= l*l-(l-1);                            //lmax是这个层的中间那个数
        int dis = n - lmid;
        //分情况讨论
        if (l % 2==0 && dis >= 0) printf("%d %d", l, l - dis);
        else if (l % 2==0 && dis < 0) printf("%d %d", l + dis, l);
        else if (l % 2 && dis>=0) printf("%d %d", l - dis, l);
        else if (l % 2 && dis < 0) printf("%d %d", l, l + dis);
        printf("
");
    }
    return 0;
}