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 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

题目大意：

给你一个足够大的棋盘，有一只蚂蚁按照一定的方式走，问你在时间t，它的坐标是多少。

走的方式是，一上，一右，一下，一右，二上，二左……如蛇形盘绕。

如果我们从对角线看呢，1 3 7 13 21

公差是1 2 4 6 8 首项为1，公差为2的等差数列的前n项和。

首先找出对应的位置，用Lower_bound（）找出的是位置pos，但是可能的位置是pos,pos-1，判断是这两个位置的哪一个，然后分奇数偶数处理。

#include<iostream>
#include<string>
#include<math.h>

using namespace std;

int main() {
	int n;
	while(cin >> n) {
		if(n == 0) {
			break;		
		}	
		int c = (int)ceil(sqrt(n));
		int pos = (c-1)*c+1;
		int step; // 距离差	
		step = n - pos;
		if( (c+1)%2 == 0) { //如果是奇数的话记录下 n和pos的距离差
			if(step > 0) {
				cout<<c-step<<" "<<c<<endl;	
			}
			else {
				cout<<c<<" "<<c+step<<endl;	
			}
		}
		else {
			if(step > 0) {
				cout<<c<<" "<<c-step<<endl;
			}
			else {
				cout<<c+step<<" "<<c<<endl;	
			}
		}
	}
	return 0;
}