----------------------------------------------

#include <iostream.h>
#include <math.h>

#define MAX_N 1000
#define MAX_P 15000

int ns[MAX_N];
int N, K;
int M;

int prost[MAX_P];
int count=0;


int main()
{
    M=0;
    cin>>N;

    int i;
    for(i=0; i<N; i++)
    {
        cin>>ns[i];
        if (ns[i]>M)
            M=ns[i];
    }

    prost[0]=2;
    prost[1]=3;
    prost[2]=5;

    count=3;

    int p=7;
    while (count<M)
    {
        int j;
        int qp = (int)sqrt(p)+1;
        for (j=0; (j<count)&&(p%prost[j])&&
(qp>=prost[j]); j++);
        if (p%prost[j])
            prost[count++]=p;

        p+=2;
    }

    for (i=0; i<N; i++)
        cout<<prost[ns[i]-1]<<endl;

    return 0;
}

You tried to solve problem 1069. Your solution on C++ was
compiled with the following errors:

69333
temp\69333(39) : error C2668: 'sqrt' : ambiguous call to
overloaded function


Wonder how I got it ? Just check "I want to recieve result
by email" And you'll get it.

I suggest using something like sqrt((double)p)
It is a common errror