【计算几何】The Queen’s Super-circular Patio
The Queen’s Super-circular Patio
题目描述
ring and the central stone. The stones in the other rings will touch the two adjacent stones in the next inner ring and their neighbors in the same ring. The fi gures below depict a patio with one ring of three stones and a patio with 5 rings of 11 stones. The patio is to be surrounded by a fence that goes around the outermost stones and straight between them (the heavier line in the fi gures).

The queen does not yet know how many stones there will be in each circle nor how many circles of stones there will be. To be prepared for whatever she decides, write a program to calculate the sizes of the stones in each circle and the length of the surrounding fence. The radius of the central stone is to be one queenly foot.
输入
Each data set consists of a single line of input. It contains the data set number, K, the number, N (3 ≤ N ≤ 20), of stones in each circle and the number, M (1 ≤ M ≤ 15), of circles of stones around the central stone.
输出
样例输入
3
1 3 1
2 7 3
3 11 5
样例输出
1 6.464 79.400
2 3.834 77.760
3 2.916 82.481
【参考博客】
https://www.cnblogs.com/starve/p/11164586.html
【队友代码】
#include<bits/stdc++.h>
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <cstring>
#include <string>
#include <queue>
#include <deque>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <utility>
#include <set>
#include <bitset>
#include <vector>
#define pi acos(-1.0)
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3fLL
#define ms(a,b) memset(a,b,sizeof(a))
typedef long long ll;
const int maxn=1e4+;
double r[];
int main()
{
int t;
int x;
int k,m;
scanf("%d",&t);
while(t--)
{
memset(r,,sizeof r);
scanf("%d",&x);
printf("%d ",x);
scanf("%d%d",&k,&m);
double s=pi/k;
r[]=1.0*sin(s)/(1.0-sin(s)); double b=-(2.0*((+r[])/tan(s)+r[]));
double a=(1.0/(tan(s)*tan(s)));
double c=*r[]+;
r[]=(-b+sqrt(b*b-4.0*a*c))/(2.0*a); for(int i=;i<=m+;i++)r[i]=(r[i-]*r[i-])*1.0/r[i-]*1.0;
double ans=(*k+*pi)*r[m+];
printf("%.3f %.3f\n",r[m+],ans); }
return ;
}
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