太简单了

勾股定理：

代码：

#include<iostream>
#include<cmath>
#define QwQ return 0;
using namespace std;
int a,b,c;
int main(){
	cin>>a>>b>>c;
	if(a>b)swap(a,b);
	if(b>c)swap(b,c);
	if(a>b)swap(a,b);
	if(a*a+b*b==c*c)cout<<a*b/2;
	else cout<<"no";
	QwQ
}

呵呵，不想手写排序直接三个元素sort（

#include<bits/stdc++.h>
#define int long long
#define INF 0x3f3f3f
using namespace std;
int a[5];
signed main(){
	cin>>a[1]>>a[2]>>a[3];
	sort(a+1,a+4);
	if(a[1]*a[1]+a[2]*a[2]==a[3]*a[3])
		cout<<a[1]*a[2]/2;
	else cout<<"no";
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
signed main(){
	int a[4];
	cin>>a[1]>>a[2]>>a[3];
	sort(a+1,a+4);
	if(a[1]*a[1]+a[2]*a[2]!=a[3]*a[3])cout<<"no";//勾股定理
	else cout<<a[1]*a[2]/2;
}

和前面某道题差不多，也是勾股定理。这次我写个数组版本：

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int a[3];
	cin>>a[0]>>a[1]>>a[2];
	sort(a,a+3);
	if(a[0]*a[0]+a[1]*a[1]==a[2]*a[2])
	{
		cout<<a[0]*a[1]/2;
	}
	else
	{
		cout<<"no";
	}
	return 0;
}

C++ :

#include<bits/stdc++.h>
using namespace std;
int main(){
	double a;
	double b;
	double c;
	cin>>a;
	cin>>b;
	cin>>c;
	if (a*a+b*b==c*c){
		cout<<round(a*b/2);
	}
	else if(a*a+c*c==b*b){
		cout<<round(a*c/2);
	}
	else if(b*b+c*c==a*a){
		cout<<round(b*c/2);
	}
	else{
		cout<<"no";
	}
	return 0;
}

Python :

# coding=utf-8
a=int(input())
b=int(input())
c=int(input())
if a*a+b*b==c*c:
    print(int(a*b/2))
elif a*a+c*c==b*b:
    print(int(a*c/2))
elif b*b+c*c==a*a:
    print(int(b*c/2))
else:
    print('no')