Codeforces Round #100 A. New Year Table
2 seconds
256 megabytes
standard input
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.
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.
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.
4 10 4
YES
5 10 4
NO
1 10 10
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
- 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 ...
- 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& ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- 梦回----32位CPU和64位CPU的通用寄存器
1 32位Intel的CPU通用寄存器 32位CPU所含有的寄存器有:4个数据寄存器(EAX.EBX.ECX和EDX):2个变址和指针寄存器(ESI和EDI):2个指针寄存器(ESP和EBP):6个段 ...
- GPIO——端口位设置/清除寄存器BSRR,端口位清除寄存器BRR
端口位设置/复位寄存器BSRR: 注:如果同时设置了BSy和BRy的对应位,BSy位起作用. 位31:16 BRy: 清除端口x的位y (y = 0…15) 这些位只能写入并只能以字(16 ...
- CSU 1258 异或运算的线段树
题目大意:在给定区间内对每个数的最后一个二进制为1的位将其修改为0,如果数本身已经为0了,就不做改变 输出给定区间的所有数的异或值 #include <cstdio> #include & ...
- 【bzoj3505】[Cqoi2014]数三角形
[bzoj3505][Cqoi2014]数三角形 2014年5月15日3,5230 Description 给定一个nxm的网格,请计算三点都在格点上的三角形共有多少个.下图为4×4的网格上的一个三角 ...
- sql 日期问题从周转换到日期
alter procedure p_date@year int=2005, --年份@week int=33, --第几周@firstday datetime =null output, ...
- Quartz进一步学习与使用
一.再思考 了解Quartz.NET的基本使用方法了.但如果想方便的知道某个作业执行情况,需要暂停,启动等操作行为,这时候就需要个Job管理的界面,如何才能达到我们想到的效果,查看相关Quartz.n ...
- 【BZOJ3110】K大数查询(权值线段树套线段树+标记永久化,整体二分)
题意:有N个位置,M个操作.操作有两种,每次操作 如果是1 a b c的形式表示在第a个位置到第b个位置,每个位置加入一个数c 如果是2 a b c形式,表示询问从第a个位置到第b个位置,第C大的数是 ...
- 主席树初探--BZOJ2588: Spoj 10628. Count on a tree
n<=100000的点权树,有m<=100000个询问,每次问两个点间的第k小点权,保证有解,强制在线. 主席上树啦!类似于之前的序列不带修改询问的前缀表示法,现在只要把前缀当成某点到根的 ...
- Swoole 入门学习(二)
Swoole 入门学习 swoole 之 定时器 循环触发:swoole_timer_tick (和js的setintval类似) 参数1:int $after_time_ms 指定时间[毫秒] ...
- ArrayAdapter的使用
package com.pingyijinren.test; import android.content.Context; import android.view.LayoutInflater; i ...