题解

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

double m(double a,double b,double c){
    double p=(a+b+c)/2;
    double x=sqrt(p*(p-a)*(p-b)*(p-c));//计算n*(n-a)*(n-b)*(n-c)的平方根 
    return x;
}

int main(){
    double a,b,c;
    cin>>a>>b>>c;
    double n=m(a,b,c);
    printf("%.2f",n);
	
    return 0;
}