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 nR 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
4 10 4
output
YES
input
5 10 4
output
NO
input
1 10 10
output
YES

本人喜欢用余弦定理...

推公式,选两个相邻的小圆的圆心与大圆圆心连线。然后2*pi/n就是这个角的最小值,然后余弦定理求这个角对应的边的长度与2*r相比

注意精度  1e-9

/* ***********************************************
Author :guanjun
Created Time :2016/7/26 11:34:41
File Name :cf100a.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << ;
const double eps=1e-;
#define pi 4.0*atan(1.0)
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
int x,y;
};
struct cmp{
bool operator()(Node a,Node b){
if(a.x==b.x) return a.y> b.y;
return a.x>b.x;
}
}; bool cmp(int a,int b){
return a>b;
}
int main()
{
#ifndef ONLINE_JUDGE
//freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
double n,R,r;
while(cin>>n>>R>>r){
if(n==){
if(r<=R)puts("YES");
else puts("NO");
continue;
}
double tmp=(pi/n);
double c=sin(tmp)*(R-r);
if(r<=c+eps){
puts("YES");
}
else puts("NO"); }
return ;
}

Codeforces Round #100 A. New Year Table的更多相关文章

  1. 贪心 Codeforces Round #273 (Div. 2) C. Table Decorations

    题目传送门 /* 贪心:排序后,当a[3] > 2 * (a[1] + a[2]), 可以最多的2个,其他的都是1个,ggr,ggb, ggr... ans = a[1] + a[2]; 或先2 ...

  2. Codeforces Round #144 (Div. 2) D table

    CodeForces - 233D 题目大意给你一个n*m 的矩阵,要求你进行涂色,保证每个n*n的矩阵内都有k个点被涂色. 问你一共有多少种涂色方案. n<=100 && m& ...

  3. Codeforces Round #345 (Div. 1) C. Table Compression dp+并查集

    题目链接: http://codeforces.com/problemset/problem/650/C C. Table Compression time limit per test4 secon ...

  4. Codeforces Round #345 (Div. 2) E. Table Compression 并查集

    E. Table Compression 题目连接: http://www.codeforces.com/contest/651/problem/E Description Little Petya ...

  5. Codeforces Round #273 (Div. 2)-C. Table Decorations

    http://codeforces.com/contest/478/problem/C C. Table Decorations time limit per test 1 second memory ...

  6. codeforces Codeforces Round #345 (Div. 1) C. Table Compression 排序+并查集

    C. Table Compression Little Petya is now fond of data compression algorithms. He has already studied ...

  7. Codeforces Round #100(140~~)

    140 A. New Year Table 题目大意:有一个大圆桌子,半径是R, 然后有n个半径是r的盘子,现在需要把这些盘子摆放在桌子上,并且只能摆放在桌子边缘,但是不能超出桌子的范围....问能放 ...

  8. codeforces 的 Codeforces Round #273 (Div. 2) --C Table Decorations

    C. Table Decorations time limit per test 1 second memory limit per test 256 megabytes input standard ...

  9. Codeforces Round #273 (Div. 2)C. Table Decorations 数学

    C. Table Decorations   You have r red, g green and b blue balloons. To decorate a single table for t ...

随机推荐

  1. 笔试算法题(28):删除乱序链表中的重复项 & 找出已经排好序的两个数组中的相同项

    出题:给定一个乱序链表,节点值为ASCII字符,但是其中有重复项,要求去除重复项并保证不改变剩余项的原有顺序: 分析:创建一个256(2^8)大小的bool数组,初始化为false,顺序读取链表,将字 ...

  2. buf.values()

    buf.values() 返回:{Iterator} 创建并返回一个包含 Buffer 值(字节)的迭代器.当 Buffer 使用 for..of 声明时将自动调用该函数. const buf = B ...

  3. Django 模版语法 一

    创建项目 django_template 和 app django-admin startproject django_template python manage.py startapp app01 ...

  4. PyQt5+requests实现车票查询工具

    PyQt5+requests实现一个车票查询工具,供大家参考,具体内容如下 结构图   效果图   思路 1.search(QPushButton)点击信号(clicked)连接到自定义的槽函数(ev ...

  5. selenium切换frame(iframe)

    例如网页代码为: <!DOCTYPE html><html lang="en"><head> <meta charset="UT ...

  6. SSM java.lang.NullPointerException

    如上图所示的异常 原来是少了这个东西,加上就好了

  7. 九度oj 题目1192:回文字符串

    题目1192:回文字符串 时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:4391 解决:2082 题目描述: 给出一个长度不超过1000的字符串,判断它是不是回文(顺读,逆读均相同)的. ...

  8. nyoj 96 n-1位数(处理前导 0 的情况)(string)

    n-1位数 时间限制:3000 ms  |            内存限制:65535 KB 难度:1   描述 已知w是一个大于10但不大于1000000的无符号整数,若w是n(n≥2)位的整数,则 ...

  9. hdu 3697 贪心

    #include<stdio.h> #include<stdlib.h> #include<string.h> #define inf 0x3fffffff #de ...

  10. bzoj 2653 middle (可持久化线段树)

    middle Time Limit: 20 Sec  Memory Limit: 512 MBSubmit: 1981  Solved: 1097[Submit][Status][Discuss] D ...