UVa 10161 Ant on a Chessboard (简单数学)
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10161 - Ant on a Chessboard
Time limit: 3.000 seconds
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=99&page=show_problem&problem=1102
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
题意:找到数字对应的行和列。
思路:找规律,奇数行,起始为行数的平方。偶数列,起始为列数的平方。行和列有与数字匹配的规律。依次确定所在的行和列。
完整代码:
/*0.012s*/#include<cstdio>#include<cmath>int main(void){int sqr, n, x, y, corner;while (scanf("%d", &n), n){sqr = (int)sqrt(n - 1) + 1;corner = sqr * sqr - sqr + 1;if ((sqr & 1) == 0){if (n >= corner){x = sqr;y = sqr - (n - corner);}else{x = sqr - (corner - n);y = sqr;}}else{if (n >= corner){x = sqr - (n - corner);y = sqr;}else{x = sqr;y = sqr - (corner - n);}}printf("%d %d\n", x, y);}return 0;}
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