CodeForces 141B - Hopscotch

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本文深入探讨了游戏开发中嵌入式硬件（如51单片机、stm32）与软件（如Unity3D、Unreal Engine）的集成应用，详细阐述了如何将硬件资源有效整合到游戏开发流程中，实现高性能的游戏体验。

So nearly half of the winter is over and Maria is dreaming about summer. She's fed up with skates and sleds, she was dreaming about Hopscotch all night long. It's a very popular children's game. The game field, the court, looks as is shown in the figure (all blocks are square and are numbered from bottom to top, blocks in the same row are numbered from left to right). Let us describe the hopscotch with numbers that denote the number of squares in the row, staring from the lowest one: 1-1-2-1-2-1-2-(1-2)..., where then the period is repeated (1-2).

$\text{[math]}$

The coordinate system is defined as shown in the figure. Side of all the squares are equal and have length a.

Maria is a very smart and clever girl, and she is concerned with quite serious issues: if she throws a stone into a point with coordinates(x, y), then will she hit some square? If the answer is positive, you are also required to determine the number of the square.

It is believed that the stone has fallen into the square if it is located strictly inside it. In other words a stone that has fallen on the square border is not considered a to hit a square.

Input

The only input line contains three integers: a, x, y, where a (1 ≤ a ≤ 100) is the side of the square, x and y ( - 106 ≤ x ≤ 106, 0 ≤ y ≤ 106) are coordinates of the stone.

Output

Print the number of the square, inside which the stone fell. If the stone is on a border of some stone or outside the court, print "-1" without the quotes.

Sample test(s)

input
1 0 0

output
-1

input
3 1 1

output
1

input
3 0 10

output
5

input
3 0 7

output
-1

input
3 4 0

output
-1

===============================

除了1号格子以外其他的都是1-2循环

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>

using namespace std;

int main()
{
    int a,x,y,ans;
    cin>>a>>x>>y;
    if(x>=a||x<=-a||y%a==0||y<=0) cout<<"-1"<<endl;
    else if(x<(a+1)/2&&x>-((a+1)/2)&&y<a) cout<<"1"<<endl;
    else
    {
        y-=a;
        if((y/a+1)%2)
        {
            if(x<(a+1)/2&&x>-((a+1)/2))
            {
                ans=y/a/2*3+2;
                cout<<ans<<endl;
            }
            else cout<<"-1"<<endl;
        }
        else
        {
            if(x>0)
            {
                ans=(y/a+1)/2*3+1;
                cout<<ans<<endl;
            }
            else if(x<0)
            {
                ans=(y/a+1)/2*3;
                cout<<ans<<endl;
            }
            else cout<<"-1"<<endl;
        }
    }
    return 0;
}