Codeforces Round #652 (Div. 2) #A. FashionabLee

A.FashionabLee

time limit per test2 seconds memory limit per test256 megabytes input:standard input output:standard output

Lee is going to fashionably decorate his house for a party, using some regular convex polygons.

Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.

Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.

Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.

Input

The first line contains a single integer t ($1\le t\le 10^4$) — the number of polygons in the market.

Each of the next t lines contains a single integer ni ($3\le n_i\le 10^9$): it means that the i-th polygon is a regular $n_i$-sided polygon.

Output

For each polygon, print ‘YES’ if it’s beautiful or NO otherwise (case insensitive).

Example

• Input:
  4
  3
  4
  12
  1000000000

• Output:
  NO
  YES
  YES
  YES

  该题题意为给出正多边形的边数，判断该正多边形是否可以满足有一边平行于x轴，一边平行于y轴，即满足有两边互相垂直即可。
  这是一个纯粹的数学问题，通过求解正多边形的每一个内角和对应的外角$\theta$，若满足外角$\theta$恰好能被90°整除，则满足条件，但是注意到外角不都为整数，比如边为11时得到的外角$\theta$为32.72°，此时无法用取模来判断是否可以被90°整除。
  于是注意到外角$\theta$可以被90°整除时，自然也可以被180°整除，通过第一条平行于x轴的边可以分别得到两条垂直于x轴的边，两条平行于x轴的边，易知这四条边恰好关于正多边形为中心对称，根据对称性，当且仅当正多边形的边数n为4的倍数时得解。
代码如下：

#include
using namespace std;
void solve();
int main()
{
	int t;
	cin>>t;
	while(t--)
		solve();
	return 0;
}
void solve()
{
	int n;
	cin>>n;
	if(n%4==0)cout<<"YES"<<endl;
	else cout<<"NO"<<endl;
}

  这道题的代码很简单，但是解题思路才是学习的最重要的部分，有了数学思维，学习程序才能设计出更加高效的代码。