Codeforces Round #100 A. New Year Table （几何精度）

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题目链接：http://codeforces.com/problemset/problem/140/A

题意：给你一个大圆半径R，一个小圆半径r，问是否能在R中放置n个小圆，小圆必须贴着大圆的边

思路：求出r在R中最多能放多少个，注意特判，向下取整过程中需要加精度

ac代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<set>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 1010000#define LL long long#define ll __int64#define INF 0xfffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1)#define eps 1e-8using namespace std;ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}//headint main(){    double n,R,r;    int cnt;    while(scanf("%lf%lf%lf",&n,&R,&r)!=EOF)    {        if(r>R||r*r*n>R*R)        {            printf("NO\n");            continue;        }        if(r*2.0>R)            cnt=1;        else if(r*2.0==R)            cnt=2;        else        {            cnt=0;            double l=R-r;            double j=(l*l-2.0*r*r)/(l*l);            double x=acos(j);            cnt+=(int)floor((2.0*PI)/x+eps);        }        //printf("%d\n",cnt);        if(cnt<n)            printf("NO\n");        else            printf("YES\n");    }    return 0;}

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