题目地址：点这里
思路：可以先确定A，，B的坐标，然后再通过确定向量来硬算出角度。。好像可以推公式做，没推出来╮(╯_╰)╭
AC代码：
#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <cmath>using namespace std;const double PI = 4 * atan(1.0);struct Point {double x, y;Point(double x = 0, double y = 0) : x(x) , y(y) { } };typedef Point Vector; Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); }Vector operator – (Vector A, Vector B) { return Vector(A.x-B.x, A.y-B.y); }Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); } bool operator < (const Point& a, const Point& b) {return a.x < b.x || (a.x == b.x && a.y < b.y);} const double eps = 1e-10;int dcmp(double x) {if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1;}bool operator == (const Point& a, const Point& b) {return dcmp(a.x – b.x) == 0 && dcmp(a.y – b.y) == 0;}double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; } double Length(Vector A) { return sqrt(Dot(A, A)); }double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); } double Cross(Vector A, Vector B) { return A.x*B.y – A.y*B.x; }double Area2(Point A, Point B, Point C) { return Cross(B-A, C-A); }Vector Rotate(Vector A, double rad) {return Vector(A.x*cos(rad) – A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad) );} Vector Normal(Vector A) {double L = Length(A);return Vector(-A.y/L, A.x/L); }Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {Vector u = P – Q;double t = Cross(w, u) / Cross(v, w);return P + v * t;} double DistanceToLine(Point P, Point A, Point B) {Vector v1 = B-A, v2 = P – A;return fabs(Cross(v1,v2) / Length(v1)); } double DistanceToSegment(Point P, Point A, Point B) {if(A==B) return Length(P-A);Vector v1 = B – A, v2 = P – A, v3 = P – B;if(dcmp(Dot(v1, v2)) < 0) return Length(v2);else if(dcmp(Dot(v1, v3)) > 0) return Length(v3);else return fabs(Cross(v1, v2)) / Length(v1); } Point GetLineProjection(Point P, Point A, Point B) {Vector v = B – A;return A + v * ( Dot(v, P-A) / Dot(v, v) ); } bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {double c1 = Cross(a2 – a1, b1 – a1), c2 = Cross(a2 – a1, b2 – a1),c3 = Cross(b2 – b1, a1 – b1), c4 = Cross(b2 – b1, a2 – b1);return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;} bool OnSegment(Point p, Point a1, Point a2) {return dcmp(Cross(a1 – p, a2 – p)) == 0 && dcmp(Dot(a1 – p, a2 – p)) < 0;} double ConvexPolygonArea(Point* p, int n) {double area = 0;for(int i = 1; i < n-1; i++)area += Cross(p[i] – p[0], p[i + 1] – p[0]);return area / 2; }int main() {double a, b, c, d, e;while(scanf("%lf %lf %lf %lf %lf", &a, &b, &c, &d, &e) != EOF && a) {if(dcmp(a + b + c + d + e – 180) != 0) {printf("Impossible\n"); continue;}a = a / 180 * PI;b = b / 180 * PI;c = c / 180 * PI;d = d / 180 * PI;e = e / 180 * PI;Point A = Point(0,0);Point B = Point(1,0);Vector AC = Vector(1,tan(b+c));Vector BC = Vector(1,-tan(d+e));Vector AE = Vector(1,tan(c));Vector BD = Vector(1,-tan(e));Point E = GetLineIntersection(A, AE, B, BC);Point D = GetLineIntersection(A, AC, B, BD);double ans = Angle(D-E, A-E);printf("%.2lf\n", ans / PI * 180); }return 0;}
真正的强者，不是流泪的人，而是含泪奔跑的人。
