someone said about for(1..999^2), there is it.
so nowtime i think - is there some tests which will be failed with this solution? on TIMUS tests it's got AC...

#include <stdio.h>

#define SUM 1000
#define DELTA 1

int main()
{
    float f1,f2,f3, i,j, k1,k2,k3, t1,t2,t3, m,mx=0;

    scanf("%f%f%f",&k1,&k2,&k3);

    for(f1=0;f1<=100;f1++)
        for(f2=0;f2<=100-f1;f2++)
        {
            f3=(100-f1-f2);
            t1 = SUM * (f1/100) * k1;
            t2 = SUM * (f2/100) * k2;
            t3 = SUM * (f3/100) * k3;
            m=t1;
            if(t2<m) m=t2;
            if(t3<m) m=t3;
            if(m>mx)
            {
                mx=m;
                i=f1; j=f2;
            }
        }

    for(f1=i-DELTA;f1<=i+DELTA;f1+=0.001)
        for(f2=j-DELTA;f2<=j+DELTA;f2+=0.001)
        {
            f3=(100-f1-f2);
            t1 = SUM * (f1/100) * k1;
            t2 = SUM * (f2/100) * k2;
            t3 = SUM * (f3/100) * k3;
            m=t1;
            if(t2<m) m=t2;
            if(t3<m) m=t3;
            if(m>mx) mx=m;
        }

    printf("%0.0f",mx);
    return 0;
}

Edited by author 16.12.2007 21:04

And with binary search ;)

bool good(double k[3], double x0)
{
  double x1 = k[0] / k[1] * x0;
  double x2 = k[0] / k[2] * x0;

  return x0 + x1 + x2 <= 1000.0;
}

int main()
{
  double k[3];
  for (int i = 0; i < 3; i++) cin >> k[i];

  sort(k, k + 3);

  double l = 0, r = 1000;
  while (r - l >= 0.001)
  {
    double m = l + (r - l) / 2;

    if (good(k, m))
      l = m;
    else
      r = m;
  }

  cout << int(l * k[0] + 0.5);
}