牛客挑战赛38

A 多边形与圆

题意

任意凸多边形在圆内滚动，由1号点开始离开到1号点再次滚动到圆内1号点所滚动的弧长是多少？

思路

计算几何

$\angle k1=acos\frac{x[i][i-1]}{2R}\quad \angle k2=acos\frac{x[i][i+1]}{2R} \\ \angle i=acos\frac{x[i][i-1]^2+x[i][i+1]^2-x[i-1][i+1]^2}{2*x[i][i-1]*x[i][i+1]} \\ \angle \theta=\angle k1+\angle k2-\angle i \\l=\angle \theta *x[1][i]$
acos返回值为弧度制。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
const double eps = 1e-8;
const int NINF = 0xc0c0c0c0;
const int INF  = 0x3f3f3f3f;
const ll  mod  = 1e9 + 7;
const ll  maxn = 1e6 + 5;
const int N = 200;

ll n,R;
double x[N][N],a[N],b[N],d[N];

double len(int i,int j){
	return sqrt((a[i]-a[j])*(a[i]-a[j])+(b[i]-b[j])*(b[i]-b[j]));
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>n>>R;
	for(int i=1;i<=n;i++){
		cin>>a[i]>>b[i];
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			x[i][j]=len(i,j);
		}
	}
	double res=0;
	for(int i=2;i<=n;i++){
		double u=acos(x[i][i-1]/(2.0*R));
		double v=acos(x[i][i%n+1]/(2.0*R));
		double w=acos(((x[i][i-1])*(x[i][i-1])+(x[i][i%n+1])*(x[i][i%n+1])-(x[i%n+1][i-1])*(x[i%n+1][i-1]))/(2.0*x[i][i-1]*x[i][i%n+1]));
		double k=(u+v-w)*x[1][i];
		res+=k;
	}
	printf("%.10lf",res);
	return 0;
}