#include <stdio.h>
#include <math.h>

double k1, k2, k3;

int main() {
    scanf("%lf%lf%lf", &k1, &k2, &k3);
    printf("%.0lf", 1000/(k2/k3+k2/k1+1)*k2);
    return 0;
}

WHY "CE" IN GCC(linux) it works.
#include <iostream>
#include <math.h>
double a,b,c,r;
using namespace std;
int main() {
    cin>>a>>b>>c;
    r=1000.0/(1.0/b+1.0/c+1.0/a);
    cout<<int(round(r));
    return 0;
}

Because here's compiler don't know function round, you must write your own function, and you can use printf() it will round the number

Edited by author 19.08.2010 11:47

#include <iostream>
using namespace std;
#include <cmath>
void main()
{
    double k1,k2,k3;
    cin>>k1>>k2>>k3;
    cout.setf(ios::fixed);
    cout.precision(0);
    cout<<1000.0*k1*k2*k3/(k1*k2+k1*k3+k2*k3);
}

#include <iomanip>

cout<<fixed<<setprecision(0)<<s;

ruby


a1,a2,a3 = gets.chomp!.split(" ").map(&:to_f)
puts (1000*a2/(a2/a1+a2/a3+1)).round