CF 171（div2）A

螺旋路径上的转向次数计算

最新推荐文章于 2023-01-28 20:01:07 发布

原创最新推荐文章于 2023-01-28 20:01:07 发布 · 783 阅读

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本文介绍了一种算法，用于计算一匹名为Valera的马沿着平面坐标系上的无限螺旋路径从起点(0, 0)移动到目标点(x, y)时所需的转向次数。螺旋由一系列整数坐标段构成，覆盖了平面上的所有整数点。

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 andy (|x|, |y| ≤ 100).

Output

Print a single integer, showing how many times Valera has to turn.

Sample test(s)

Input

0 0

Output

Input

1 0

Output

Input

0 1

Output

Input

-1 -1

Output

直接模拟，一步步走过去，每一步走的是转角的点，分四种情况走，对于那些不处在转角的终点的处理方式，，就是分四种情况判断它是在哪一段上终止。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#define LL long long
using namespace std;
const int maxn = 1000000 + 5;

int main(){
    int x,y;
    while(cin >> x >> y){
        if(x == 0 && y == 0) {
            cout << 0 << endl;
            continue;
        }
        int px = 0,py = 0;
        int ans = 0;
        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++;
            }
        }
        cout << ans-1 << endl;
    }
    return 0;
}