2 seconds
256 megabytes
standard input
standard output
Valera the horse lives on a plane. The Cartesian coordinate system is defined on this plane. Also an infinite spiral is painted on the plane. The spiral consists of segments: [(0, 0), (1, 0)], [(1, 0), (1, 1)], [(1, 1), ( - 1, 1)], [( - 1, 1), ( - 1, - 1)], [( - 1, - 1), (2, - 1)],[(2, - 1), (2, 2)] and
so on. Thus, this infinite spiral passes through each integer point of the plane.
Valera the horse lives on the plane at coordinates (0, 0). He wants to walk along the spiral to point (x, y).
Valera the horse has four legs, so he finds turning very difficult. Count how many times he will have to turn if he goes along a spiral from point (0, 0) to
point(x, y).
The first line contains two space-separated integers x and y (|x|, |y| ≤ 100).
Print a single integer, showing how many times Valera has to turn.
0 0
0
1 0
0
0 1
2
-1 -1
3
解题说明：此题是可以看成是一个螺旋形迷宫，从最中心的位置（0,0）走到指定位置（x，y）。为了求解需要转弯的次数，可以模拟人在里面走动，遇到拐角了转弯次数加1，然后判断终点是不是在转过去的那段直线上，如果是就结束，否则继续走下去，直到走到终点所在的那段直线上。当然，如果用几何的方式求解速度更快，不过暂时还没找出几何规律。
#include<iostream>
#include<map>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
int x,y;
int px = 0,py = 0;
int ans = 0;
scanf("%d %d",&x,&y);
if(x == 0 && y == 0)
{
printf("0\n");
}
else
{
while(1)
{
if(px == x && py == y)
{
break;
}
else if(px == x && x > 0 && y < py && y > (-py+1))
{
break;
}
else if(py == y && y > 0 && x > px && x < -px)
{
break;
}
else if(px == x && x < 0 && y > py && y < -py)
{
break;
}
else if(py == y && y < 0 && x < px && x > (-px+1))
{
break;
}
if(px > 0 && py > 0)
{
px = -px;
ans++;
}
else if(px < 0 && py > 0)
{
py = -py;
ans++;
}
else if(px <= 0 && py <= 0)
{
px = -px + 1;
ans++;
}
else
{
py = -py + 1;
ans++;
}
}
printf("%d\n",ans-1);
}
return 0;
}