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.
3
30
60
90
NO
YES
YES
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 }
