Edited by author 21.10.2021 22:47

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int main()
{
    long long val = 0;
    vector<double> v;
    while (cin >> val &&val!=EOF)
    {
        v.push_back(sqrt(val));
    }
    for (int i = v.size()-1; i >=0; i--)
    {
        cout<< v[i] << endl;
    }
    return 0;
}

Some AC to try this test

BID 0.01
BID 10000
BID 5000
BID 5000
SALE 5000 3
DEL 5000
SALE 3000 3
SALE 0.01 3
QUIT

Hint: ->, not <->

Rot13: Gur qenhtug bayl zbirf sbejneq jura chfurq, abg ovqverpgvbanyyl.

the reason is min(ti+(ti*T+99)/100,100500*ti) can overflow.
So you are supposed to change long long into unsigned long long,then you'll get AC.

I think that this is too easy problem for 297 points of hardness

This case may help you:
2
A B C
B D Isenbaev

ans:
A 2
B 1
C 2
D 1
Isenbaev 0

Change the statement, don't be ashamed

Try this:
3 3
2 2
3 3

answer: 4

I wish they made clarification

My friends solve it with dicnic.
But its time complexity can be up to O(nm^sqrt(n+m)),and its space compexity can be up to O(nm).
I think admin should add the test that all the bank belong to one country.

I see many people had this WA. Can someone give me a hint about it?
I think it will be useful for others.

AMOGUS

#include <iostream>
using namespace std;

main()
{
    int a, b ;
    int a1 , b1;
    int a2 , b2;
    cin >> a >> b;
    cin >> a1 >> b1;
    cin >> a2 >> b2;

    int c = b - b1;
    int c1 = a - a2;
    cout << c1 << " " <<  c;
    return 0;
}

____________________________________
Держите)

import java.util.Scanner;
public class Main
{
    public static void main(String[] args) {
        Scanner a = new Scanner(System.in);
        int x = a.nextInt();
        //String x1 = a.nextLine();
        int y = 0;
        int z = 0;
        int [] array1 = new int[x];
        for(int i=0;i<x;i++)
        {
            array1[i] = a.nextInt();
            y+=array1[i];

            if(i==1 && array1[i] < array1[i-1] )
                {
                    int tmp = array1[i];
                    array1[i] = array1[i-1];
                    array1[i-1] = tmp;
                }
        }

        for(int i=array1.length-1;i>-1;i--)
        {
            z+=array1[i]*y;
            y-=array1[i];
            z+=array1[i]*y;
        }

        System.out.println(z);
    }
}

#include <iostream>
int main() {
    short n;int a=0;
    std::cin>>n;
    if(n<=0){
        for(int i=n;i<=1;i++){
            a+=i;
        }
        std::cout<<a;
    }
    else{
        for(int i=1;i<=n;i++){
            a+=i;
        }
        std::cout<<a;
    }
return 0;
}

The answer is the symmetry axis
In test 3 the angle between the lines is very small

Edited by author 26.06.2015 02:34

Почему так не проходит(ошибка на 13):
#include <iostream>
int main() {
    int n,k;
    std::cin>>n>>k;
    if(n<=k) std::cout<<2;
    else std::cout<<n*2/k+n*2%k;

return 0;
}

А так проходит:
    #include <iostream>
int main() {
    int n,k;
    std::cin>>n>>k;
    if(n<=k) std::cout<<2;
    else{
        if(n*2 % k == 0) std::cout<<n*2/k;
        else std::cout<<n*2/k+1;
    }
return 0;
}
Ну или тернар(тоже не проходит - ошибка на 13):
#include <iostream>
int main() {
    int n,k;
    std::cin>>n>>k;
    (n<=k)?std::cout<<2:std::cout<<n*2/k+n*2 % k;

return 0;
}
Может я чего не понимаю, может всё-таки где-то приоритет потерял, но ведь идентично же.

Edited by author 08.10.2021 11:51
Теперь понял, %k может вернуть и больше, чем 1

Edited by author 08.10.2021 12:00

Edited by author 08.10.2021 12:02
C++ ответ:
#include <iostream>
int main() {
    int n,k;
    std::cin>>n>>k;
    (n<=k)?std::cout<<2:(n*2%k==0)?std::cout<<n*2/k:std::cout<<n*2/k+1;

return 0;
}

Edited by author 08.10.2021 12:03