codeforces 141 B. Hopscotch (规律)

B. Hopscotch

time limit per test
 2 seconds

memory limit per test
 256 megabytes

input
 standard input

output
 standard output

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).



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.

Examples

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

题意：像小时候玩的跳格子，在一个直角坐标上，这里会有很多层，每层1 1 2 1 2 （重复1 2），给一个点，点在方格内且不压线，输出放个的序号

否则输出-1；

code：
#include <iostream>
#include<cstdio>
using namespace std;

int main()
{
int a,x,y;
scanf("%d%d%d",&a,&x,&y);
int b=y/a;//落在哪一行
b+=1;
if(y%a==0)
printf("-1\n");
else//行数奇偶分析，在分析x坐标
{
if(b==1)
{
if(2*x<a&&2*x>(-a))
printf("1\n");
else
printf("-1\n");
}
else if(b%2==0)
{
if(2*x<a&&2*x>(-a))
printf("%d\n",(b-2)/2*3+2);
else
printf("-1\n");
}
else if(b%2)
{
if(x<0&&x>(-a))
printf("%d\n",(b-1)/2*3);
else if(x>0&&x<a)
printf("%d\n",(b-1)/2*3+1);
else
printf("-1\n");
}
}
return 0;
}