Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.

He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle a.

Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to a?

The first line of input contains an integer t (0?<?t?<?180) — the number of tests. Each of the following t lines contains a single integer a(0?<?a?<?180) — the angle the robot can make corners at measured in degrees.

For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.

In the first test case, it is impossible to build the fence, since there is no regular polygon with angle .

In the second test case, the fence is a regular triangle, and in the last test case — a square.

 题解：看是否能构成正多边形

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <vector>
 6 #include <cstdlib>
 7 #include <iomanip>
 8 #include <cmath>
 9 #include <ctime>
10 #include <map>
11 #include <set>
12 using namespace std;
13 #define lowbit(x) (x&(-x))
14 #define max(x,y) (x>y?x:y)
15 #define min(x,y) (x<y?x:y)
16 #define MAX 100000000000000000
17 #define MOD 1000000007
18 #define pi acos(-1.0)
19 #define ei exp(1)
20 #define PI 3.141592653589793238462
21 #define INF 0x3f3f3f3f3f
22 #define mem(a) (memset(a,0,sizeof(a)))
23 typedef long long ll;
24 const int N=1005;
25 const int mod=1e9+7;
26 int main()
27 {
28     std::ios::sync_with_stdio(false);
29     int t;
30     cin>>t;
31     while(t--){
32         int n;
33         cin>>n;
34         if(360%(180-n)==0) cout<<"YES"<<endl;
35         else cout<<"NO"<<endl;
36     }
37     return 0;
38 }