10161 - Ant on a Chessboard

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于 2013-08-01 00:23:13 发布

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分类专栏： Volume 1. Elementary Problem 文章标签：刘汝佳算法人们经典 AOAPC I Maths - Misc

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本文介绍了一种计算在特定规则下，蚂蚁在棋盘上行走至任意给定时间位置的方法，通过数学规律和算法实现路径预测。

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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

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

Sample Output

2 3

5 4

1 5

找规律：副对角线之间差值成等差数列，且根据给出的n可求出所在的最小方格数m，于是可求出对角线的值=(m-1)^2-m，在根据m的奇偶性，左、下的变化规律即可求得。

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
    int n,r,c,m,dia;
    while(cin>>n&&n)
    {
        m=sqrt((double)n);
        if(m*m<n)
            m++;
        r=c=m;
        dia=(m-1)*(m-1)+m;
        if(m%2==0)
        {
            if(dia>=n)
                c-=dia-n;
            else
                r-=n-dia;
        }
        else
        {
            if(dia>=n)
                r-=dia-n;
            else
                c-=n-dia;
        }
        cout<<c<<' '<<r<<endl;
    }
    return 0;
}