2018 Multi-University Training Contest 5 E.Everything Has Changed（余弦定理）

Edward is a worker for Aluminum Cyclic Machinery. His work is operating mechanical arms to cut out designed models. Here is a brief introduction of his work.
Assume the operating plane as a two-dimensional coordinate system. At first, there is a disc with center coordinates (0,0) and radius R. Then, m mechanical arms will cut and erase everything within its area of influence simultaneously, the i-th area of which is a circle with center coordinates (xi,yi) and radius ri (i=1,2,⋯,m). In order to obtain considerable models, it is guaranteed that every two cutting areas have no intersection and no cutting area contains the whole disc.
Your task is to determine the perimeter of the remaining area of the disc excluding internal perimeter.
Here is an illustration of the sample, in which the red curve is counted but the green curve is not.

问大圆切掉m个小圆后的周长（如图红色部分）。

构造一个三边分别为R、r、d（圆心距）的三角形，就可以使用余弦定理，求出大圆和小圆半个圆心角的余弦值，再使用arccos，就能求出这一段的弧度，最后用弧长公式，对弧度*r，再扩大两倍，就是要求的弧长了

大圆的周长减去切掉的弧长，加上增加的弧长就是答案。

ac代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const double PI = 3.14159265358979;

int main() {
    int t, m, R;
    double dis, arc, arc2, ans;
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d", &m, &R);
        ans = 2 * PI * R;
        while(m--) {
            int x, y, r;
            scanf("%d%d%d", &x, &y, &r);
            dis = x * x + y * y;
            // 跳过内含
            if(dis < (R - r) * (R - r)) {
                continue;
            }
            // 内切直接加小圆周长，相交的则计算两段弧长
            if(dis == (R - r) * (R - r)) {
                ans += 2 * PI * r;
            } else if(dis < (R + r) * (R + r)) {
                dis = sqrt(dis);
                arc = acos((r * r + dis * dis - R * R) / (2 * r * dis)) * r * 2;
                arc2 = acos((R * R + dis * dis - r * r) / (2 * R * dis)) * R * 2;
                ans += arc;
                ans -= arc2;
            }
        }
        // 这题实际上对精度要求非常低
        printf("%.20f\n", ans);
    }
    return 0;
}