Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Description
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).
Input
The first line contains two space-separated integers x and y(|x|, |y| ≤ 100).
Output
Print a single integer, showing how many times Valera has to turn.
Sample Input
0 0
0
1 0
0
0 1
2
-1 -1
3 题意：给出一个从原点开始螺旋前进的移动规则，现要到达有限范围内的某个坐标，问一共需要几次转向。
模拟。
思路：画一下图找下规律就OK了，开始我以为输入的点只是该点所在正方形上的顶点，其实它是该正方形上任意一个整数点。
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std; int main()
{
int x,y;
while(~scanf("%d %d",&x,&y))
{
if(x == && y == )
{
printf("0\n");
continue;
}
int k = max(abs(x),abs(y));
int ans = (k-)*;
if(x >= k- && x <= k && y == -k)
printf("%d\n",ans);
else if(x == k && y >= -k && y <= k)
printf("%d\n",ans+);
else if(x >= -k && x <= k && y == k)
printf("%d\n",ans+);
else if(x == -k && y >= -k && y <= k)
printf("%d\n",ans+);
else if(x >= -k && x <= k && y == -k)
printf("%d\n",ans+);
}
return ;
}