#include <iostream>;
#include <vector>;

using namespace std;

int main()
{
    vector<int> v;
    int n;
    while (std::cin >> n)
        v.push_back(n);

    int i;
    for(i = v.size() - 1; i>=0; i--)
        printf("%.4f\n", sqrt((double) (v[i])));

    system("pause");
    return 0;
}

P.S.
А тесты на которых проверяются программы можно увидеть?

I cant find my mistake please help me what is test 5.

Edited by author 19.10.2013 17:12

What a bad problem it is!
It is more like a reading comprehension problem in an English exam than a programming problem!

Что означает Problem locked?

Do the segments include their endpoints?

what's wrong in my code:
#include <iostream>
int main()
{
    int N;
    std::cin >> N;
    int result=0;
    if (N=0){
        std::cout << 1 << std::endl;
        return 0;
    }
    if (N > 0)
    {
        for (int i=2; i<=N;i++)
        {
            result+=i;
        }
        std::cout << result << std::endl;
        return 0;
    }else
    {
        for (int i=-2;i >=N;i--)
        {
            result+=i;
        }
        std::cout << result << std::endl;
        return 0;
    }
    return 0;
}

Edited by author 18.10.2013 23:00

How to deal with it?

I saw many guys got AC very fast,is there any way to work the problem out other than just find-cycle?

There is a catch, you should account for three situations:
1) N > 0
2) N < 0
3) N == 0 <--- 8-D.

Small assignment, but took an hour of trying.

I think that code is easy to understand.
If you got WA, try it.

#include <iostream>
#include <map>
using namespace std;

int main() {
    map<char, string> m;

    int n; cin >> n;
    while (n--) {
        char c1, c2; cin >> c1 >> c2;
        m[c1].push_back(c2);
    }
    char c; cin >> c;
    for (int i = 0; i < m[c].size(); ++i)
        cout << c << m[c][i] << endl;

    return 0;
}

#include <iostream>
using namespace std;
int main ()
{
    short int i, n1, n2, n3;
    cin >> i;
    cin.sync();
    char *name1, *name2, *name3, *name  = new char[20];
    n1 = n2 = n3 = 0;
    name1 = "Emperor Penguin";
    name2 = "Macaroni Penguin";
    name3 = "Little Penguin";
    while (i > 0)
    {
        cin.getline(name,19);
        cin.sync();
        switch(name[0])
        {
        case 'E': n1++;
            break;
        case 'M': n2++;
            break;
        case 'L': n3++;
            break;
        }
        i--;
    }
    if (n1 > n2 && n1 > n3)
        printf("%s",name1);
    else
        if (n2 > n1 && n2 > n3)
        printf("%s",name2);
        else
            if(n3 > n1 && n3 > n2)
            printf("%s",name3);
    system("pause");
    return 0;
}

used STL multimap and scanf,printf instead cin,cout
got AC 0.5 sec ;)

Must add length to the final result.

Solve it with brute force, afaik this problem is of NP class.

import java.util.*;

public class My {

    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);
        int a = in.nextInt();

        if ((a >= 1) && (a <= 4)){
            System.out.println("few");
        } else if ((a >= 5) && (a <= 9)){
            System.out.println("several");
        } else if ((a >= 10) && (a <= 19)){
            System.out.println("pack");
        } else if ((a >= 20) && (a <= 49)){
            System.out.println("lots");
        } else if ((a >= 50) && (a <= 99)){
            System.out.println("horde");
        } else if ((a >= 100) && (a <= 249)){
            System.out.println("throng");
        } else if ((a >= 250) && (a <= 499)){
            System.out.println("swarm");
        } else if ((a >= 500) && (a <= 999)){
            System.out.println("zounds");
        } else if (a > 1000){
            System.out.println("legion");
        }

    }

}

Please help me!!!

i guess the right way is like
AAB
AAB
BBC

Edited by author 25.01.2013 16:44

Brute force (полный перебор) method gives Time Limit Exceeded. In Java it takes 6 seconds to for parameters N=8, K=10. So the clever method (which uses recurrent calls) is the way to go.