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  • Codeforce 270A

    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?

    Input

    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.

    Output

    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.

    Examples
    input
    3
    30
    60
    90
    output
    NO
    YES
    YES
    Note

    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 }
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  • 原文地址:https://www.cnblogs.com/wydxry/p/7262358.html
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