CodeForces 610APasha and Stick(数学)

Description
Pasha has a wooden stick of some positive integer length n. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be n.

Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square.

Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer x, such that the number of parts of length x in the first way differ from the number of parts of length x in the second way.

Input
The first line of the input contains a positive integer n (1 ≤ n ≤ 2·109) — the length of Pasha's stick.

Output
The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square.

Sample Input
Input
6
Output
1
Input
20
Output

4

代码：

#include
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
	if(n%2==1)
	printf("0\n");
	else
	{
	if(n%4==0)
	printf("%d\n",n/4-1);
	else
	printf("%d\n",n/4);	
	}
	}
	return 0;
}

思路：题目为给你一个数n，问将n分成4份组成多少个矩形而且不是正方形。

首先判断是否为偶数，不是偶数肯定不可以。矩形对边相等，所以可以看成将n分成两个偶数，这样就有n/4种可能。然后判断是不是4的倍数，是的话答案要减去1个，不是的话直接输出（这道水题也wa了一次，没有判断是不是偶数，题目没说我就想当然的以为给的都是偶数还是太傻）