这题就设AB是1，，然后正弦余弦定理去搞搞搞就可以了
代码：
#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const double PI = acos(-1.0);int a, b, c, d, e;double get(double x) {return x / 180.0 * PI;}int main() {while (~scanf("%d%d%d%d%d", &a, &b, &c, &d, &e) && a || b || c || d || e) {if (a + b + c + d + e != 180) {printf("Impossible\n");continue;}double ACB = a, CAE = b, EAB = c, CBD = d, DBA = e;double AB = 1.0;double AOB = 180 – EAB – DBA;double CBA = CBD + DBA;double CAB = EAB + CAE;double AEB = 180 – EAB – CBA;double ADB = 180 – DBA – CAB;double BD = AB / sin(get(ADB)) * sin(get(CAB));double AE = AB / sin(get(AEB)) * sin(get(CBA));double AO = AB / sin(get(AOB)) * sin(get(DBA));double BO = AB / sin(get(AOB)) * sin(get(EAB));double EO = AE – AO;double DO = BD – BO;double DE = sqrt(EO * EO + DO * DO – DO * EO * 2 * cos(get(AOB)));double x = asin(DO * sin(get(AOB)) / DE) * 180 / PI;printf("%.2f\n", x);}return 0;}
空虚无聊的时候就读书，但一定得有自己的生活目标和计划。
