Uva 10161 - Ant on a Chessboard
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
#include<stdlib.h> #include<string.h> #include<math.h> int main() { int x, y, n, flag, m; while(scanf("%d", &n) == 1 && n) { m = (int)sqrt((double)n); if(m*m < n) m++; flag = n - (m*m-m+1); if(m%2 == 0) { if(flag>=0) { x = m; y = m - flag; } else { x = m + flag; y = m; } } else { if(flag>=0) { x = m - flag; y = m; } else { x = m; y = m + flag; } } printf("%d %d\n", x, y); } return 0; }
解题思路:
找规律,用线将数字按顺序圈出来再分析一下就简单了很多
说实话, 一个月没碰题了,找条简单的水题YY希望能够允许吧?
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