jrMz and angles（水题）

jrMz and angles

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1071    Accepted Submission(s): 560

Problem Description

jrMz has two types of angles, one type of angle is an interior angle of n-sided regular polygon, and the other type of angle is an interior angle of m-sided regular polygon. jrMz wants to use them to make up an angle of 360 degrees, which means, jrMz needs to choose some or none of the angles which belong to the first type and some or none of the angles which belong to the second type so that the sum value of the angles chosen equals 360 degrees. But jrMz doesn’t know whether it is possible, can you help him?

 

Input

The first line contains an integer T(1≤T≤10)——The number of the test cases.
For each test case, the only line contains two integers n,m(1≤n,m≤100) with a white space separated.

 

Output

For each test case, the only line contains a integer that is the answer.

 

Sample Input

3 4 8 3 10 5 8

 

Sample Output

Yes Yes No

Hint

In test case 1, jrMz can choose 1 angle which belongs to the first type and 2 angles which belong to the second type, because 90+135+135=360. In test case 2, jrMz can choose 6 angles which belong to the first type, because6 imes60=360. In test case 3, jrMz can’t make up an angle of 360 degrees.

题解：给两个正n,m边形，选角度凑成360度，每个角度是(n - 2)*180/n

代码：

import java.util.Scanner;

public class Main {
    //(n - 2) * 180/n
    private static Scanner cin;
    static{
        cin = new Scanner(System.in);
    }
    
    static double getAng(int x){
        return (x - 2.0) * 180.0 / x;
    }
    
    static boolean isEqual(double x1, double x2){
        if(Math.abs(x1 - x2) < 1e-15){
            return true;
        }else{
            return false;
        }
    }
    static boolean work(int n, int m){
        double x1 = getAng(n);
        double x2 = getAng(m);
        for(int i = 0; i * x1 <= 360.0; i++){
            for(int j = 0; i * x1 + j * x2 <= 360.0;j++){
                double temp = i * x1 + j * x2;
                if(isEqual(temp, 360)){
                    return true;
                }
            }
        }
        return false;
    }
    public static void main(String[] args) {
        int T;
        T = cin.nextInt();
        while(T-- > 0){
            int n = cin.nextInt();
            int m = cin.nextInt();
            if(work(n, m)){
                System.out.println("Yes");
            }else{
                System.out.println("No");
            }
        }
    }
}

A. New Year Table

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.

Input

The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.

Output

Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".

Remember, that each plate must touch the edge of the table.

Examples

input

output

input

output

input

output

Note

The possible arrangement of the plates for the first sample is:

 

题意：

给定一个大圆的半径 R 以及 n 个小院的半径 r 。

问能否把这 n 个小圆贴着大圆边缘放下。

思路：

2*PI/(2 * a)就好了；注意精度要加上1e-15

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const double PI = acos(-1.0);

int main(){
    int n;
    double R, r;
    while(~scanf("%d%lf%lf", &n, &R, &r)){
        if(n == 1){
            if(R >= r){
                puts("YES");
            }else{
                puts("NO");
            }
            continue;
        }
        if(R == r){
            if(n <= 1)
                puts("YES");
            else
                puts("NO");
            continue; 
        }
        double a = asin(1.0 * r/(R - r));
        double cnt = 2.0 * PI / (2 * a);
        cnt += 1e-15;
        if(n <= cnt){
            puts("YES");
        }else{
            puts("NO");
        }
    }
    return 0;
}