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最新推荐文章于 2025-07-06 16:11:00 发布

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本文探讨如何通过给定的奇数n和整数m，在由1到n*n构成的螺旋矩阵中找到对应的行号和列号。算法通过计算对角线数值、确定所在层以及计算与目标值的距离来定位目标值的位置。

Problem Description

Given an odd number n, we can arrange integers from 1 to n*n in the shape of a spiral. The figure below illustrates the spiral made by integers from 1 to 25.

1 2 3 4 5

------------------------

1 | 21 22 23 24 25

2 | 20 7 8 9 10

3 | 19 6 1 2 11

4 | 18 5 4 3 12

5 | 17 16 15 14 13

As we see above, each position in the spiral corresponds to a unique integer. For example, the number in row 1, column 1 is 21, and integer 16 is in row 5, column 2.
Now, given the odd number n (1<=n<=32768), and an integer m (1<=m<=n*n), you should write a program to find out the position of m.

Input

The first line of the input is a positive integer T(T<=20). T is the number of the test cases followed. Each case consists of two integer n and m as described above.

Output

For each case, output the row number and column number that the given integer is in, separated by a single whitespace. Please note that the row and column number are both starting from 1.

Sample Input

Sample Output

1 3
1 1
5 2

Author

HYNU

代码：

/*题目数值大，要推理出来，找下标规律
先把对角线一半的数值推出来，并保存
再确定在哪一层之后，算出距离，
*/
#include<cstdio>
#include<iostream>
using namespace std;
int a[32769];
int main()
{
    int i, j, t, m, n;
    int count = 0, x, y;
    while(scanf("%d", &t) != EOF)
    {
        while(t--)
        {

            scanf("%d%d", &n, &m);
            count = 0;
            for(i = 0; i < n; i++) //得出对角线的数值
                if(i == 0) a[i] = 1;
                else a[i] = 8 * i - 2 + a[i - 1];

            while(a[count] < m)  count++; //确定在哪一层

            i = j = n / 2 + 1 - count; //确定行列坐标的大概位置

            x = a[count] - m; //距离 左下半部分(竖行) 的距离(从行列到具体数值的距离)
            if(x <= count * 4)
            {
                if(x <= count * 2) //列
                    i = i + x;
                else//行
                {
                    i = i + count * 2;
                    j = j + x - count * 2;
                }
            }
            else
            {
                y = x - count * 4; //距离 右上半部分(竖行) 的距离(从行列到具体数值的距离)
                if(y <= count * 2 - 1) //列
                {
                    i = i + count * 2 - y;
                    j = j + count * 2;
                }
                else//行(比本行的对角数还要大)
                {
                    y = y - (count * 2 - 1);
                    i = i + 1;
                    j = j + count * 2 - y;
                }
            }
            printf("%d %d\n", i, j);
        }
    }
    return 0;
}