def j(): a,b,c=map(float,input('请输入三角形三条边的长度,用空格隔开:').split()) if a>0 and b>0 and c>0 and a+b>c and a+c>b and b+c>a: l=a+b+c p=l/2 s=p*(p-a)*(p-b)*(p-c)#海伦公式 print('三角形的周长:{:.2f}\n三角形的面积:{:.2f}'.format(l,s)) else: print('三角形不成立,请重新输入') j…
思路:海伦公式, AC代码: #include<bits/stdc++.h> using namespace std; int main() { int n; scanf("%d",&n); double ha, hb, hc, a, b, c; while(~scanf("%lf %lf %lf",&ha,&hb,&hc)) { a = 2.0 / ha; b = 2.0 / hb ; c = 2.0 / hc; if(…