题目：点击打开链接
题意： 给你一个原点在(0,0)，半径为R的圆盘以及m个小圆。第i个小圆的原点在(xi,yi)，半径为ri。每个小圆可能会与大圆盘相交，现在问你m个小圆将大圆覆盖后的周长并。（题目保证小圆不会相交，且小圆不会完全把大圆覆盖）
分析：余弦定理，大大简化运算。
代码：
#pragma comment(linker, "/STACK:102400000,102400000")
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<string>
#include<cstdio>
#include<bitset>
#include<vector>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<deque>
#include<list>
#include<set>
#include<map>
using namespace std;
#define debug test
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
#define eps 1e-10
#define PI acos(-1.0)
typedef pair<int,int> PII;
const ll mod = 1e9+7;
const int N = 1e6+10;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int to[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int t;
int main() {
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
scanf("%d", &t);
while(t--) {
int n, R;
scanf("%d%d", &n, &R);
double ans = 0, rem = PI;
for(int i = 1; i <= n; ++i) {
int x, y, r;
scanf("%d%d%d", &x, &y, &r);
int dis2 = x * x + y * y;
if(dis2 < (R - r) * (R - r) || dis2 > (R + r) * (R + r))
continue;
double dis = sqrtl(dis2);
double ang1 = acos((dis2 + R * R - r * r) / (2 * R * dis));
double ang2 = acos((dis2 - R * R + r * r) / (2 * r * dis));
rem -= ang1;
ans += 2 * r * ang2;
}
ans += 2 * R * rem;
printf("%.20f\n", ans);
}
return 0;
}