Простенькие задания

#include <iostream.h>
#include <iomanip.h>
#include <math.h>
main()
{
int n,i;
double max_otr;
cout<<"enter size of massive\n";
cin>>n;
double a[n];
for (i=1;i<n;i++)
a[i]=rand()%n-20;
for (i=1;i<n;i++)
{
if (a[i]>0) continue;
else 
{
max_otr=abs(a[i]);
break;
}
}
for (i=2;i<n;i++)
{
if (a[i]<0 && abs(a[i])>max_otr)
max_otr=abs(a[i]);
else continue;
}
cout<<"max otricateln="<<max_otr<<"\n";
return 0;
}

 #include<iostream>
#include<conio.h>
#include<math.h>
Using namespace std;
int main()
{
Double a,b,c;
сout<<введите a,b,c<<endl;
cout<<a=;
сin>>a;
cout<<b=;
сin>>b;
cout<<c=;
сin>>c;
double sqrt(p(p-a)(c-b)(p-c) где р=с/2
сout<<S=<S;
return 0;
getch();
}

main()
{
...
P1=P/2;
S=sqrt(P1*(P1-a)*(P1-b)*(P1-c));
...
}

#include <iostream>
#include <math.h>
using namespace std;
main()
{
    float a,b,c,C,S,p;
 
    cout << "Enter a side";
    cin >> a;
    cout << "Enter b side";
    cin >> b;
    cout << "Enter c side";
    cin >> c;
 
    C = a+b+c;
    p = C/2;
 
    
    S =  sqrt(p(p-a)*(p-b)*(p-c));
    cout << "C =" << C << "; " << "S =" << S;
    return 0;
}

a[i]=rand()%n-20