Codeforces 112B-Petya and Square（实现）

B. Petya and Square

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Petya loves playing with squares. Mum bought him a square 2n × 2n in size. Petya marked a cell inside the square and now he is solving the
following task.

The task is to draw a broken line that would go along the grid lines and that would cut the square into two equal parts. The cutting line should not have any common points with the marked cell and the resulting two parts should be equal up
to rotation.

Petya wants to determine whether it is possible to cut the square in the required manner given the sizes of the square side and the coordinates of the marked cell. Help him.

Input

The first line contains three space-separated integers 2n, x and y (2 ≤ 2n ≤ 100, 1 ≤ x, y ≤ 2n),
representing the length of a square's side and the coordinates of the marked cell. It is guaranteed that 2n is even.

The coordinates of the marked cell are represented by a pair of numbers x y,
where x represents the number of the row and y represents
the number of the column. The rows and columns are numbered by consecutive integers from 1 to 2n.
The rows are numbered from top to bottom and the columns are numbered from the left to the right.

Output

If the square is possible to cut, print "YES", otherwise print "NO"
(without the quotes).

Sample test(s)

input

4 1 1

output

YES

input

2 2 2

output

NO

题意 ：有一块2n*2n的地板，在地板上有一个东西，给出坐标和地板的边长。如今要将地板切成相等的两部分，且不能切到那个东西，问能不能切成功。

仅仅要在中间一定不能切。

。反之则可

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <set>
#include <map>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
using namespace std;
const int INF = 0x3f3f3f3f;
#define LL long long
char s[1000000];
int main()
{
	int n,x,y;
	while(~scanf("%d%d%d",&n,&x,&y))
	{
		int tx=n/2,ty=n/2;
		if((x==tx||x==tx+1)&&(y==ty||y==ty+1))
			puts("NO");
		else
			puts("YES");
	}

	return 0;
}