Скажите, пожалуйста, как можно на Вашем сайте создать свою "олимпиаду", турнир по программированию.

Please Help and give me some tests!

Edited by author 01.11.2017 20:58

If in the line "outputon", this does not mean that you need to put "out" and "puton", since sometimes you can put "output" and "one".

Сheck out this test:
1
outputone
Answer : "YES"

Perhaps this will help you.

Edited by author 01.11.2017 18:28

Edited by author 01.11.2017 18:29

 If start from all the leaf airport that from one airport could win, start from this airport suerly will lose. Otherwise,he could win.

My program had such mistake: it has printed results in the order in which athletes have finished the race, but the task is to print results in order in which athletes have started one.

Here is my code which has been written in java. Please help me. I cannot understand what wrong is. Or some tests you have ?. Thanks.

public class EsreveRredrO_1226 {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String s = reader.readLine();
        solve(s);
    }

    public static void solve(String s) {
        String text[];
        text = s.split(" ");
        for (int i = 0; i < text.length; i++) {
            String revers = "";
            String temp;
            for (int j = text[i].length() - 1; j >= 0;  j--) {
                    revers = revers.concat(String.valueOf(text[i].charAt(j)));
            }
            if (revers.charAt(0) == '!' || revers.charAt(0) == ',' || revers.charAt(0) == '.' || revers.charAt(0) == '?'
                    || revers.charAt(0) == ':' || revers.charAt(0) == ' ') {
                temp = revers.substring(1, revers.length());
                revers = temp.concat(String.valueOf(revers.charAt(0)));
            }
            if (i != text.length - 1)
            System.out.print(revers + " ");
            else System.out.print(revers);
        }
    }
}

Edited by author 04.09.2017 00:45

6
1 3 2 6 5 4

answer: 0

Edited by author 31.10.2017 02:47

it's one of the most awful descriptions of the problems i've ever seen. some corner cases, which are really questionable, are not explained. it really irritates because instead of solving the next problem you need to guess what the answer is. how should i know what is the answer for 1? i am not linguist, but smth tells me, that there are no situation when you can find elements in between in the list of one element. what is the answer? should i put 0, 1, etc? after all it's not a competition, i can't even ask the judge or stuff how should i interpret the problem itself.

Edited by author 31.10.2017 02:27

I wrote simple implementation but can't figure out test to prove my solution wrong.

4 3 -> 4
3 3 -> 2
4 5 -> 6
4 6 -> 7
3 8 -> 8
6 5 -> 8
6 8 -> 10

#include "stdio.h"
#include <stdlib.h>

int main(void) {
  int bigger, lesser, a, b;
  scanf("%d", &a);
  scanf("%d", &b);
  if (a >= b) {
    bigger = a - 1; lesser = b - 1;
  } else {
    bigger = b - 1; lesser = a - 1;
  }

  if (bigger % lesser == 0) {
    printf("%d\n", bigger); return 0;
  }

  printf("%d\n", bigger + lesser - 1);

  return 0;
}

4 100
aaaaaaaaaaaaaaaaaaaa11aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa111
ans : 4

Edited by author 15.10.2006 11:24

346 158 618
955.0 105.0
891.0 93.0

65.1152, isn't it?????

Edited by author 21.04.2008 16:38

#include <iostream>
using namespace std;
int main()
{
    int n;
    cin>>n;
    if(1<=n&&n<=12)
    {
        int a=12-n;
    if(a*45<=4*60)
    cout<<"YES";
    else
    cout<<"NO";
    }
return 0;
}

Tell me please how you have get timing 0.001 ? It mustn't be a code, may be just method or algorithm. Thanks

If I divide first and then add 1 with (n or other group numbers), I get WA. But if I add 1 at first and then I divide that time i get AC.
I have seen some of the discussion, but couldn't find the logic.Can someone help me out?

#include <iostream>
#include <iomanip>
#include <map>
#include <vector>
#include <cmath>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;

int letter(char c)
{
    return (('a'<=c&&c<='z')||('A'<=c&&c<='Z'));
}

int main()
{
    //freopen("a","r",stdin);
    string s;
    while(getline(cin,s))
    {
        vector<int> l,r;
        for (size_t i=0;i<s.length();i++)
        {
            if (letter(s[i]))
            {
                if (!i||!letter(s[i-1]))
                {
                    l.push_back(i);
                }
                if (i+1==s.length()||!letter(s[i+1]))
                {
                    r.push_back(i+1);
                }
            }
        }
        for (int i=0;i<l.size();i++)
        {
            reverse(s.begin()+l[i],s.begin()+r[i]);
        }
        cout << s << endl;
    }
    return 0;
}

Edited by author 28.10.2017 19:36

I test all case in discussion,but all answers right...
Can someone help..Thanks!!!

I know how to construct larger cases when knowing solution n<=11&&m<=11,but brute_force for small tests is tooooooooo slow..