p822和p879的结合，想进阶可做p825

#include<iostream>
#include<iomanip>
#include<stdio.h>
#include<math.h>
#include<string>
#include<cstring>
using namespace std;
int main(){
    double a,b,c,p,maxx,sum;
	cin>>a>>b>>c;
	maxx=max(a,max(b,c));
	sum=a+b+c;
	if(maxx<sum-maxx){
		p=(a+b+c)/2;
	    printf("%.2lf",sqrt(p*(p-a)*(p-b)*(p-c)));
	}else{
		cout<<"No Solution.";
	}
}

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{
	float a,b,c,ans,p;
	cin >> a >> b >> c;
	if (a + b > c && a + c > b && b + c > a)
	{
		p = (a + b + c) / 2;
		ans = sqrt(p * (p - a) * (p - b) * (p - c));
		cout << fixed << setprecision(2) << ans;
	}
	else
		cout << "No Solution.";
	return 0;
}

#include <iostream>
#include <algorithm>//sort()头文件
#include <cmath>//sqrt()头文件
using namespace std;
float a[4], p;//输出要求浮点数，故可直接用float定义，为方便排序，可直接用数组存储a, b, c
int main()
{
	cin >> a[1] >> a[2] >> a[3];
	sort(a + 1, a + 4);
	if(a[1] + a[2] <= a[3])
	{
		puts("No Solution.");
		return 0;
	}
	p = (a[1] + a[2] + a[3]) / 2;
	printf("%.2f\n", sqrt(p * (p - a[1]) * (p - a[2]) * (p - a[3])));
	return 0;
}

#include <stdio.h>
#include <iostream>
#include <math.h>
#include<iomanip>
using namespace std;
int main()
{
	double a,b,c,p,x;
	cin >> a >> b >> c;
	p=(a+b+c)/2;
	x=(p-a)*(p-b)*(p-c)*p;
	double s=sqrt(x);
	if(a+b > c && a + c > b && c + b > a)
	{
		cout<<fixed<<setprecision(2)<<s<<endl;
	}
		else
		{
			cout<<"No Solution.";
		}

}

#include <stdio.h> #include #include <math.h> #include using namespace std; int main() { double a,b,c,p,x; cin>>a>>b>>c; p=(a+b+c)/2; x=(p-a)(p-b)(p-c)*p; double s=sqrt(x); if(a+b>c&&a+c>b&&c+b>a) { cout<<fixed<<setprecision(2)<<s<<endl; } else { cout<<"No Solution."; }

}