codeforces 270A Fancy Fence

最新推荐文章于 2024-10-30 02:03:01 发布

风吹过的烟花

最新推荐文章于 2024-10-30 02:03:01 发布

阅读量306

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM算法_水题

本文链接：https://blog.youkuaiyun.com/zhang__liuchen/article/details/73251686

ACM算法_水题专栏收录该内容

76 篇文章

订阅专栏

本文探讨了如何使用一台仅能在特定角度构建拐角的机器人来建造一个理想的正多边形围栏。通过数学分析，我们得出结论：如果360度能被(180度-指定角度)整除，则机器人可以成功构建所需的围栏。

网址：点击打开链接

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 $\text{[math]}$ .

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

分析：利用正多边形的一个角的补角，判断是否符合条件，正多边形的角的补角和为360°。

#include<stdio.h>
#include<map>
#include<string>
using namespace std;
int main()
{
    int t,a;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&a);
        if(360%(180-a)==0)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}