这题就海伦公式带进去就可以了。。
要注意的是，这题的样例，，是输入n次错误的输入才停止。。，输入的可能是负数。
代码：
#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const double eps = 1e-8;int t;double Ha, Hb, Hc;int dcmp(double x) {if (fabs(x) < eps) return 0;else return x < 0 ? -1 : 1;}int main() {scanf("%d", &t);while (t) {scanf("%lf%lf%lf", &Ha, &Hb, &Hc);double a = 2 / Ha, b = 2 / Hb, c = 2 / Hc;double p = (a + b + c) / 2;double tmp = p * (p – a) * (p – b) * (p – c);if (dcmp(Ha) <= 0 || dcmp(Hb) <= 0 || dcmp(Hc) <= 0 || dcmp(tmp) <= 0) {t–;printf("These are invalid inputs!\n");continue;}double ans = 1.0 / sqrt(tmp);printf("%.3f\n", ans);}return 0;}
上帝助自助者。
