dp[i][j] - можно ли получить остаток j если длина равна i.
я получал всякие превышение памяти и времени, только потому что я сохранял предка. Но если не сохранять предка, а вычислять самому, то решение будет < 100 миллисекунд. Восстанавливал и находил минимальный ответ через рекурсию  почти как дфс.

Don't use sqrt.

Use a dictionary instead of a big if-else ladder.

Cracked it after 7 years :)

Has anybody AC with Python 3.3?

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);
    }
}