Spiral——找规律暴力

Contest02-4 Spiral

Time Limit: 1000MS Memory limit: 65536K

题目描述

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



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.

输入

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.

输出

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.

示例输入

3
3 9
5 21
5 16

示例输出

1 3
1 1
5 2

提示

 题意：给出一个n * n的图，中间是1的位置，然后按照顺时针螺旋旋转，为2到n * n的位置，求任意一个数的位置坐标。

关键就是找到1的坐标，还有次对角线上的坐标，通过观察图片，可以得到：1的坐标总是n/2+1，因为n总是奇数，从1开始，沿着次对角线向右上方延伸，依次为3*3,5*5,7*7......都是奇数的平方。从1位置沿次对角线向左下方向找，依次为2*2+1,4*4+1,6*6+1.....,都是偶数的平方加1，而且，它们的坐标都可以根据1的坐标很容易地计算出来，所以以这条对角线为根本，就可以计算任何一个点的坐标了。

 

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

using namespace std;

int main()
{
int t,n,m,i,j,k,fang,x,y,x1,y1,x2,y2;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
int bj = 0;
x = y = n/2 + 1;
if(m == 1)
{
printf("%d %d\n",x,y);
}
else
{
for(i = 1; i <= 32767; i = i + 2)  //找到该点对应的圈数，即该点最后大于等于的上方对角线的位置
{
if((i + 2) *( i + 2) > m && i * i <= m)
{
x1 = x - (i - 1)/2;
y1 = y + (i - 1)/2;
break;
}
}
fang = i;
x2 = x + (i+1) /2;  //找到改圈对应的下方对角线的坐标，这样就将一个整圈分成了2个部分
y2 = y - (i+1) /2;
if(i  * i == m)
{
printf("%d %d\n",x1,y1);
continue;
}
else
{
i = i +1;
if(((fang +1)* (fang+1)) + 1 <= m)
{
x2 = x + i/2;
y2 = y - i/2;
bj = 1;
}
if(((fang +1)* (fang+1)) + 1 == m)
{
printf("%d %d\n",x2,y2);
continue;
}
else
{
if(bj == 1)  //说明点在第二圈，即图的左边或是上边
{
int flag = 0;
int aa = i * i + 1;
for(x2--; x2 >= x1-1; x2--)  //判断是不是在左边
{
aa++;
if(aa == m)
{
flag = 1;
break;
}
}
if(flag == 1)
printf(
b840
"%d %d\n",x2,y2);
else  //不在左边，那就在上边，计算下标
{
x2++;
for(y2++; y2 <= y1; y2++)
{
aa++;
if(aa == m)
{
printf("%d %d\n",x2,y2);
break;
}
}
}
}
else  //锁定该点在图的右边或者下边
{
int flag = 0;
int aa = (fang) * (fang) ;
y1++;
for(x1; x1 <= x2; x1++)  //判断是不是在图的右边
{
aa++;
if(aa == m)
{
flag = 1;
break;
}
}
if(flag == 1)
printf("%d %d\n",x1,y1);
else  //不在右边，就在下边了，计算坐标
{
x1--;
for(y1--; y1 > y2; y1--)
{
aa++;
if(aa == m)
{
printf("%d %d\n",x1,y1);
break;
}
}
}
}
}
}
}
}
return 0;
}